Linked List Merge Sort Exaplanation - algorithm

Can someone explain to me please, how this code works :
http://www.chiark.greenend.org.uk/~sgtatham/algorithms/listsort.c
I don't understand the algorithm used in this post. Thanks

Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.
Let head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so head node has to be changed if the data at original head is not the smallest value in linked list.
MergeSort(headRef)
1) If head is NULL or there is only one element in the Linked List
then return.
2) Else divide the linked list into two halves.
FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
MergeSort(a);
MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here)
and update the head pointer using headRef.
*headRef = SortedMerge(a, b);
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* function prototypes */
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL))
{
return;
}
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See http://geeksforgeeks.org/?p=3622 for details of this
function */
struct node* SortedMerge(struct node* a, struct node* b)
{
struct node* result = NULL;
/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
/* Pick either a or b, and recur */
if (a->data data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef)
{
struct node* fast;
struct node* slow;
if (source==NULL || source->next==NULL)
{
/* length next;
/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}
/* Function to print nodes in a given linked list */
void printList(struct node *node)
{
while(node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/* Function to insert a node at the beginging of the linked list */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct node* res = NULL;
struct node* a = NULL;
struct node* b = NULL;
/* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
/* Remove duplicates from linked list */
MergeSort(&a);
printf("\n Sorted Linked List is: \n");
printList(a);
getchar();
return 0;
}

Try imaging all the merges that are performed in a normal merge sort on an array: first, elements are paired up and merged into sorted subarray of length two, then these subarray of length two are paired up and merged into sorted subarray of length four and so on. Notice the length of the subarray: 1, 2, 4, and so on, let's call this instep, which doubles in each iteration.
At any point, p points to a list of length instep, q points to a list of length instep or smaller (we may hit the end of the list), and q immediately follows p. They form a pair of subarray as mentioned above. We do a merge on p and q to get a sorted list of length psize + qsize starting from p. We than move p and q to the next pair, and so on. Once we are done with the whole list, we double instep and start merging longer sorted list.

Related

Why the time complexity of the below program which converts BT to BST is nlogn (which is the time complexity of sort)?

I am confused, the function binaryTreeToBST that has the below-mentioned operations and function call. Why are we not adding time complexities of each called function instead of taking only the time complexity of sort (step 3 below)?
countNodes
Creating temp array arr[] that stores inorder traversal of the tree. (takes O(n) )
sort the temp array arr[]. This step takes O(nlogn) time.
Again do inorder traversal to convert BT to BST(takes O(n) )
/* A program to convert Binary Tree to Binary Search Tree */
#include <stdio.h>
#include <stdlib.h>
/* A binary tree node structure */
struct node {
int data;
struct node* left;
struct node* right;
};
/* A helper function that stores inorder traversal of a tree rooted
with node */
void storeInorder(struct node* node, int inorder[], int* index_ptr)
{
// Base Case
if (node == NULL)
return;
/* first store the left subtree */
storeInorder(node->left, inorder, index_ptr);
/* Copy the root's data */
inorder[*index_ptr] = node->data;
(*index_ptr)++; // increase index for next entry
/* finally store the right subtree */
storeInorder(node->right, inorder, index_ptr);
}
/* A helper function to count nodes in a Binary Tree */
int countNodes(struct node* root)
{
if (root == NULL)
return 0;
return countNodes(root->left) + countNodes(root->right) + 1;
}
// Following function is needed for library function qsort()
int compare(const void* a, const void* b)
{
return (*(int*)a - *(int*)b);
}
/* A helper function that copies contents of arr[] to Binary Tree.
This function basically does Inorder traversal of Binary Tree and
one by one copy arr[] elements to Binary Tree nodes */
void arrayToBST(int* arr, struct node* root, int* index_ptr)
{
// Base Case
if (root == NULL)
return;
/* first update the left subtree */
arrayToBST(arr, root->left, index_ptr);
/* Now update root's data and increment index */
root->data = arr[*index_ptr];
(*index_ptr)++;
/* finally update the right subtree */
arrayToBST(arr, root->right, index_ptr);
}
// This function converts a given Binary Tree to BST
void binaryTreeToBST(struct node* root)
{
// base case: tree is empty
if (root == NULL)
return;
/* Count the number of nodes in Binary Tree so that
we know the size of temporary array to be created */
int n = countNodes(root);
// Create a temp array arr[] and store inorder traversal of tree in arr[]
int* arr = new int[n];
int i = 0;
storeInorder(root, arr, &i);
// Sort the array using library function for quick sort
qsort(arr, n, sizeof(arr[0]), compare);
// Copy array elements back to Binary Tree
i = 0;
arrayToBST(arr, root, &i);
// delete dynamically allocated memory to avoid memory leak
delete[] arr;
}
/* Utility function to create a new Binary Tree node */
struct node* newNode(int data)
{
struct node* temp = new struct node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
/* Utility function to print inorder traversal of Binary Tree */
void printInorder(struct node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder(node->left);
/* then print the data of node */
printf("%d ", node->data);
/* now recur on right child */
printInorder(node->right);
}
/* Driver function to test above functions */
int main()
{
struct node* root = NULL;
/* Constructing tree given in the above figure
10
/ \
30 15
/ \
20 5 */
root = newNode(10);
root->left = newNode(30);
root->right = newNode(15);
root->left->left = newNode(20);
root->right->right = newNode(5);
// convert Binary Tree to BST
binaryTreeToBST(root);
printf("Following is Inorder Traversal of the converted BST: \n");
printInorder(root);
return 0;
}

Tree and graphs problems in Julia using struct data types, pointers and this

I want to write solve some graph/trees problems using Julia language.
Here is some good example. In C it was done this way:
Recursive C program for level order traversal of Binary Tree
#include <stdio.h>
#include <stdlib.h>
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node *left, *right;
};
/* Function prototypes */
void printGivenLevel(struct node* root, int level);
int height(struct node* node);
struct node* newNode(int data);
/* Function to print level order traversal a tree*/
void printLevelOrder(struct node* root)
{
int h = height(root);
int i;
for (i=1; i<=h; i++)
printGivenLevel(root, i);
}
/* Print nodes at a given level */
void printGivenLevel(struct node* root, int level)
{
if (root == NULL)
return;
if (level == 1)
printf("%d ", root->data);
else if (level > 1)
{
printGivenLevel(root->left, level-1);
printGivenLevel(root->right, level-1);
}
}
/* Compute the "height" of a tree -- the number of
nodes along the longest path from the root node
down to the farthest leaf node.*/
int height(struct node* node)
{
if (node==NULL)
return 0;
else
{
/* compute the height of each subtree */
int lheight = height(node->left);
int rheight = height(node->right);
/* use the larger one */
if (lheight > rheight)
return(lheight+1);
else return(rheight+1);
}
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
/* Driver program to test above functions*/
int main()
{
struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
printf("Level Order traversal of binary tree is \n");
printLevelOrder(root);
return 0;
}
I have tried to do it in Julia in similar way, but there are some problems, especially with accessing to struct elements.(like node->right and node->left).
Or some way to create Self-referential struct and function to allocate nodes.
struct node
data::Int
left::Ptr{node}
right::Ptr{node}
end
# Compute the "height" of a tree -- the number of
# nodes along the longest path from the root node
# down to the farthest leaf node.
function height = (node::Ptr{node})
if node === nothing
return 0
else
# compute the height of each subtree
lheight = height(node->left)
rheight = height(node->right)
# use the larger one
if lheight > rheight
return lheight+1
else return rheight+1
end
end
end
From what I've seen trying to recreate a problem solution in C way isn't the way, however this struct type should be useful. I just have to know how to create self-referent struct, how to allocate elements in this node and how to get them.
First of all, I would strongly suggest reading through https://docs.julialang.org/en/v1/, since Julia is different from C in quite a few ways. There are multiple things to point out here:
Per default structs in Julia are immutable, which means you cannot modify fields after it is created. It is also always passed by copy instead of by reference, since it doesn't have a specific memory address and usually gets allocated on the stack. This actually has multiple benefits for the compiler and is part of the reason why Julia can be so fast. In your example you probably want a mutable struct, which is more similar a struct in C.
In Julia, you should never have to use pointers (Ptr) directly, unless you are calling C code. Since Julia uses a garbage collector, raw pointers have a lot of gotchas when it comes to memory management and should generally just be avoided. You usually just work with objects directly or, if you want to pass immutable objects by reference, you can wrap them in Ref.
In Julia, fields are always accessed either just with a dot x.field (equivalent to getproperty(x, :field)), or in some cases getfield(x, :field). (The latter can't be overloaded by the user, which is sometimes useful). -> actually creates an anonymous function.
For your example the following should work instead:
mutable struct Node
data::Int
left::Node
right::Node
Node(data::Int) = new(data)
end
function height(node::Node, field::Symbol)
isdefined(node, field) || return 0
return height(getproperty(node, field))
end
function height(node::Node)
lheight = height(node, :left)
rheight = height(node, :right)
# use the larger one
if lheight > rheight
return lheight+1
else
return rheight+1
end
end
What the first part is doing is creating a mutable struct Node, with self-referential fields like your C example. The line Node(data::Int) = new(data) is actually an inner constructor taking just the data and if you call new directly in a mutable struct, you can leave trailing fields undefined. You can define these fields afterwards with x.field = y. If these fields are themselves mutable, you can also check if they are undefined with isdefined(x, :field). Here, I am adding another method to height, which also takes a field name, which returns the height of the field if it's defined and 0 otherwise.
You would then construct nodes and calculate their height like this:
julia> n = Node(1)
Node(1, #undef, #undef)
julia> n.left=Node(2)
Node(2, #undef, #undef)
julia> n
Node(1, Node(2, #undef, #undef), #undef)
julia> height(n)
2
Hope that helps! If you want to learn more, the documentation I linked above is usually quite good.

Implementing the Dutch National Flag Program with Linked Lists

I wanted to sort a linked list containing 0s, 1s or 2s. Now, this is clearly a variant of the Dutch National Flag Problem.
http://en.wikipedia.org/wiki/Dutch_national_flag_problem
The algorithm for the same as given in the link is:
"Have the top group grow down from the top of the array, the bottom group grow up from the bottom, and keep the middle group just above the bottom. The algorithm stores the locations just below the top group, just above the bottom, and just above the middle in three indexes. At each step, examine the element just above the middle. If it belongs to the top group, swap it with the element just below the top. If it belongs in the bottom, swap it with the element just above the bottom. If it is in the middle, leave it. Update the appropriate index. Complexity is Θ(n) moves and examinations."
And a C++ implementation given for the same is:
void threeWayPartition(int data[], int size, int low, int high) {
int p = -1;
int q = size;
for (int i = 0; i < q;) {
if (data[i] == low) {
swap(data[i], data[++p]);
++i;
} else if (data[i] >= high) {
swap(data[i], data[--q]);
} else {
++i;
}
}
}
My only question is how do we traverse back in a linked list like we are doing here in an array?
A standard singly-linked list doesn't allow you to move backwards given a linked list cell. However, you could use a doubly-linked list, where each cell stores a next and a previous pointer. That would let you navigate the list forwards and backwards.
However, for the particular problem you're trying to solve, I don't think this is necessary. One major difference between algorithms on arrays and on linked lists is that when working with linked lists, you can rearrange the cells in the list to reorder the elements in the list. Consequently, the algorithm you've detailed above - which works by changing the contents of the array - might not actually be the most elegant algorithm on linked lists.
If you are indeed working with linked lists, one possible way to solve this problem would be the following:
Create lists holding all values that are 0, 1, or 2.
Remove all cells from the linked list and distribute them into the list of elements that are equal to 0, 1, or 2.
Concatenate these three lists together.
This does no memory allocation and purely works by rearranging the linked list cells. It still runs in time Θ(n), which is another plus. Additionally, you can do this without ever having to walk backwards (i.e. this works on a singly-linked list).
I'll leave the complete implementation to you, but as an example, here's simple C++ code to distribute the linked list cells into the zero, one, and two lists:
struct Cell {
int value;
Cell* next;
}
/* Pointers to the heads of the three lists. */
Cell* lists[3] = { NULL, NULL, NULL };
/* Distribute the cells across the lists. */
while (list != NULL) {
/* Cache a pointer to the next cell in the list, since we will be
* rewiring this linked list.
*/
Cell* next = list->next;
/* Prepend this cell to the list it belongs to. */
list->next = lists[list->value];
lists[list->value] = list;
/* Advance to the next cell in the list. */
list = next;
}
Hope this helps!
As others have said, there is no way to "back up" in a linked list without reverse links. Though it's not exactly an answer to your question, the sort can be easily accomplished with three queues implementing a bucket sort with three buckets.
The advantage of queues (vice pushing on stacks) is that the sort is stable. That is, if there are data in the list nodes (other than the 0,1,2-valued keys), these will remain in the same order for each key.
This is only one of many cases where the canonical algorithm for arrays is not the best for lists.
There is a very slick, simple way to implement the queues: circularly linked lists where the first node, say p, is the tail of the queue and consequently p->next is is the head. With this, the code is concise.
#include <stdio.h>
#include <stdlib.h>
typedef struct node_s {
struct node_s *next;
int val;
int data;
} NODE;
// Add node to tail of queue q and return the new queue.
NODE *enqueue(NODE *q, NODE *node)
{
if (q) {
node->next = q->next;
q->next = node;
}
else node->next = node;
return node;
}
// Concatenate qa and qb and return the result.
NODE *cat(NODE *qa, NODE *qb)
{
NODE *head = qa->next;
qa->next = qb->next;
qb->next = head;
return qb;
}
// Sort a list where all values are 0, 1, or 2.
NODE *sort012(NODE *list)
{
NODE *next = NULL, *q[3] = { NULL, NULL, NULL};
for (NODE *p = list; p; p = next) {
next = p->next;
q[p->val] = enqueue(q[p->val], p);
}
NODE *result = cat(q[0], cat(q[1], q[2]));
// Now transform the circular queue to a simple linked list.
NODE *head = result->next;
result->next = NULL;
return head;
}
int main(void)
{
NODE *list = NULL;
int N = 100;
// Build a list of nodes for testing
for (int i = 0; i < N; ++i) {
NODE *p = malloc(sizeof(NODE));
p->val = rand() % 3;
p->data = N - i; // List ends up with data 1,2,3,..,N
p->next = list;
list = p;
}
list = sort012(list);
for (NODE *p = list; p; p = p->next)
printf("key val=%d, data=%d\n", p->val, p->data);
return 0;
}
This is now a complete simple test and it runs just fine.
This is untested. (I will try to test it if I get time.) But it ought to be at least very close to a solution.
Using a doubly linked list. If you have already implemented a linked list object and the related link list node object, and are able to traverse it in the forward direction it isn't a whole bunch more work to traverse in the reverse direction.
Assuming you have a Node object somewhat like:
public class Node
{
public Node Next;
public Object Value;
}
Then all you really need to do is change you Node class and you Insert method(s) up a little bit to keep track of of the Node that came previously:
public class Node
{
public Node Next;
public Node Previous;
public Object Value;
}
public void Insert(Node currentNode, Node insertedNode)
{
Node siblingNode = currentNode.Next;
insertedNode.Previous = currentNode;
insertedNode.Next = siblingNode;
if(siblingNode!= null)
siblingNode.previous = insertedNode;
currentNode.next = insertedNode;
}
PS Sorry, I didn't notice the edit that included the C++ stuff so it's more C#
Works for all cases by CHANGING NODES rather than NODE DATA.. Hoping its never too late!
METHOD(To throw some light on handling corner cases):
1. Keep three dummy nodes each for 0,1,2;
2. Iterate throught the list and add nodes to respective list.
3. Make the next of zero,one,two pointers as NULL.
4. Backup this last nodes of each list.
5. Now handle 8 different possible cases to join these list and Determine the HEAD.
zero one two
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
An implementation of this in C++.
Node* sortList(Node *head)
{
struct Node dummyzero,dummyone,dummytwo;
dummyzero.next = dummyone.next = dummytwo.next = NULL;
struct Node *zero =&dummyzero,*one = &dummyone,*two=&dummytwo;
Node *curr = head,*next=NULL;
while(curr)
{
next = curr->next;
if(curr->data==0)
{
zero->next = curr;
zero = zero->next;
}
else if(curr->data==1)
{
one->next = curr;
one = one->next;
}
else
{
two->next = curr;
two = two->next;
}
curr = next;
}
zero->next = one->next = two->next =NULL; //Since this dummynode, No segmentation fault here.
Node *zerolast = zero,*onelast = one,*twolast = two;
zero = dummyzero.next;
one = dummyone.next;
two = dummytwo.next;
if(zero==NULL)
{
if(one==NULL)
head = two;
else
{
head = one;
onelast->next = two;
}
}
else
{
head = zero;
if(one==NULL)
zerolast->next = two;
else
{
zerolast->next = one;
onelast->next = two;
}
}
return head;
}
The idea is to use dutch flag sorting algorithm, with a slight modification:
sort 0's and 1's as per dutch flag method,
But for 2's instead of adding them at the end of list, keep them in a separate linked list.
And finally append the 2's list to the sorted list of 0's and 1's.
Node * sort012_linked_list(Node * head) {
if (!head || !head->next)
return head;
Node * head_of_2s = NULL;
Node * prev = NULL;
Node * curr = head;
while (curr) {
if (curr->data == 0) {
if (prev == NULL || prev->data == 0) {
prev = curr;
curr = curr->next;
}
else {
prev->next = curr->next;
curr->next = head;
head = curr;
curr = prev->next;
}
}
else if (curr->data == 1) {
prev = curr;
curr = curr->next;
}
else { // curr->data == 2
if (prev == NULL) {
head = curr->next;
curr->next = head_of_2s;
head_of_2s = curr;
curr = head;
}
else {
prev->next = curr->next;
curr->next = head_of_2s;
head_of_2s = curr;
curr = prev->next;
}
}
}
if (prev)
prev->next = head_of_2s;
return head;
}

Merging 2 Binary Search Trees

How do you merge 2 Binary Search Trees in such a way that the resultant tree contains all the elements of both the trees and also maintains the BST property.
I saw the solution provided in
How to merge two BST's efficiently?
However that solution involves converting into a Double Linked List. I was wondering if there is a more elegant way of doing this which could be done in place without the conversion. I came up with the following pseudocode. Does it work for all cases? Also I am having trouble with the 3rd case.
node* merge(node* head1, node* head2) {
if (!head1)
return head2;
if (!head2)
return head1;
// Case 1.
if (head1->info > head2->info) {
node* temp = head2->right;
head2->right = NULL;
head1->left = merge(head1->left, head2);
head1 = merge(head1, temp);
return head1;
} else if (head1->info < head2->info) { // Case 2
// Similar to case 1.
} else { // Case 3
// ...
}
}
The two binary search trees (BST) cannot be merged directly during a recursive traversal.
Suppose we should merge Tree 1 and Tree 2 shown in the figure.
The recursion should reduce the merging to a simpler situation. We cannot reduce
the merging only to the respective left subtrees L1 and L2, because L2 can contain
numbers larger than 10, so we would need to include the right
subtree R1 into the process. But then we include numbers greater
than 10 and possibly greater than 20, so we would need to include
the right subtree R2 as well. A similar reasoning shows that
we cannot simplify the merging by including subtrees from Tree 1 and from Tree 2
at the same time.
The only possibility for reduction is to simplify only inside the respective trees.
So, we can transform
the trees to their right spines with sorted nodes:
Now, we can merge the two spines easily into one spine. This
spine is in fact a BST, so we could stop here. However, this BST
is completely unbalanced, so we transform it to a balanced BST.
The complexity is:
Spine 1: time = O(n1), space = O(1)
Spine 2: time = O(n2), space = O(1)
Merge: time = O(n1+n2), space = O(1)
Balance: time = O(n1+n2), space = O(1)
Total: time = O(n1+n2), space = O(1)
The complete running code is on http://ideone.com/RGBFQ. Here are the essential parts. The top level code is a follows:
Node* merge(Node* n1, Node* n2) {
Node *prev, *head1, *head2;
prev = head1 = 0; spine(n1, prev, head1);
prev = head2 = 0; spine(n2, prev, head2);
return balance(mergeSpines(head1, head2));
}
The auxiliary functions are for the tranformation to spines:
void spine(Node *p, Node *& prev, Node *& head) {
if (!p) return;
spine(p->left, prev, head);
if (prev) prev->right = p;
else head = p;
prev = p;
p->left = 0;
spine(p->right, prev, head);
}
Merging of the spines:
void advance(Node*& last, Node*& n) {
last->right = n;
last = n;
n = n->right;
}
Node* mergeSpines(Node* n1, Node* n2) {
Node head;
Node* last = &head;
while (n1 || n2) {
if (!n1) advance(last, n2);
else if (!n2) advance(last, n1);
else if (n1->info < n2->info) advance(last, n1);
else if (n1->info > n2->info) advance(last, n2);
else {
advance(last, n1);
printf("Duplicate key skipped %d \n", n2->info);
n2 = n2->right;
}
}
return head.right;
}
Balancing:
Node* balance(Node *& list, int start, int end) {
if (start > end) return NULL;
int mid = start + (end - start) / 2;
Node *leftChild = balance(list, start, mid-1);
Node *parent = list;
parent->left = leftChild;
list = list->right;
parent->right = balance(list, mid+1, end);
return parent;
}
Node* balance(Node *head) {
int size = 0;
for (Node* n = head; n; n = n->right) ++size;
return balance(head, 0, size-1);
}
Assuming we have two trees A and B we insert root of tree A into tree B and using rotations move inserted root to become new root of tree B. Next we recursively merge left and right sub-trees of trees A and B. This algorithm takes into account both trees structure but insertion still depends on how balanced target tree is. You can use this idea to merge the two trees in O(n+m) time and O(1) space.
The following implementation is due to Dzmitry Huba:
// Converts tree to sorted singly linked list and appends it
// to the head of the existing list and returns new head.
// Left pointers are used as next pointer to form singly
// linked list thus basically forming degenerate tree of
// single left oriented branch. Head of the list points
// to the node with greatest element.
static TreeNode<T> ToSortedList<T>(TreeNode<T> tree, TreeNode<T> head)
{
if (tree == null)
// Nothing to convert and append
return head;
// Do conversion using in order traversal
// Convert first left sub-tree and append it to
// existing list
head = ToSortedList(tree.Left, head);
// Append root to the list and use it as new head
tree.Left = head;
// Convert right sub-tree and append it to list
// already containing left sub-tree and root
return ToSortedList(tree.Right, tree);
}
// Merges two sorted singly linked lists into one and
// calculates the size of merged list. Merged list uses
// right pointers to form singly linked list thus forming
// degenerate tree of single right oriented branch.
// Head points to the node with smallest element.
static TreeNode<T> MergeAsSortedLists<T>(TreeNode<T> left, TreeNode<T> right, IComparer<T> comparer, out int size)
{
TreeNode<T> head = null;
size = 0;
// See merge phase of merge sort for linked lists
// with the only difference in that this implementations
// reverts the list during merge
while (left != null || right != null)
{
TreeNode<T> next;
if (left == null)
next = DetachAndAdvance(ref right);
else if (right == null)
next = DetachAndAdvance(ref left);
else
next = comparer.Compare(left.Value, right.Value) > 0
? DetachAndAdvance(ref left)
: DetachAndAdvance(ref right);
next.Right = head;
head = next;
size++;
}
return head;
}
static TreeNode<T> DetachAndAdvance<T>(ref TreeNode<T> node)
{
var tmp = node;
node = node.Left;
tmp.Left = null;
return tmp;
}
// Converts singly linked list into binary search tree
// advancing list head to next unused list node and
// returning created tree root
static TreeNode<T> ToBinarySearchTree<T>(ref TreeNode<T> head, int size)
{
if (size == 0)
// Zero sized list converts to null
return null;
TreeNode<T> root;
if (size == 1)
{
// Unit sized list converts to a node with
// left and right pointers set to null
root = head;
// Advance head to next node in list
head = head.Right;
// Left pointers were so only right needs to
// be nullified
root.Right = null;
return root;
}
var leftSize = size / 2;
var rightSize = size - leftSize - 1;
// Create left substree out of half of list nodes
var leftRoot = ToBinarySearchTree(ref head, leftSize);
// List head now points to the root of the subtree
// being created
root = head;
// Advance list head and the rest of the list will
// be used to create right subtree
head = head.Right;
// Link left subtree to the root
root.Left = leftRoot;
// Create right subtree and link it to the root
root.Right = ToBinarySearchTree(ref head, rightSize);
return root;
}
public static TreeNode<T> Merge<T>(TreeNode<T> left, TreeNode<T> right, IComparer<T> comparer)
{
Contract.Requires(comparer != null);
if (left == null || right == null)
return left ?? right;
// Convert both trees to sorted lists using original tree nodes
var leftList = ToSortedList(left, null);
var rightList = ToSortedList(right, null);
int size;
// Merge sorted lists and calculate merged list size
var list = MergeAsSortedLists(leftList, rightList, comparer, out size);
// Convert sorted list into optimal binary search tree
return ToBinarySearchTree(ref list, size);
}
The best way we could merge the trees in place is something like:
For each node n in first BST {
Go down the 2nd tree and find the appropriate place to insert n
Insert n there
}
Each iteration in the for loop is O(log n) since we are dealing with trees, and the for loop will be iterated n times, so in total we have O(n log n).
A BST is a ordered or sorted binary tree. My algorithm would be to simple :
traverse through both trees
compare the values
insert the smaller of the two into a new BST.
The python code for traversing is as follows:
def traverse_binary_tree(node, callback):
if node is None:
return
traverse_binary_tree(node.leftChild, callback)
callback(node.value)
traverse_binary_tree(node.rightChild, callback)
The cost for traversing through the BST and building a new merged BST would remain O(n)
This blog post provides a solution to the problem with O(logn) space complexity. (Pay attention that the given approach does not modify input trees.)
This can be done in 3 steps:
covert the BSTs to sorted linked list (this can be done in place with O(m+n) time)
Merge this two sorted linked lists to a single list (this can be done in place with O(m+n) time)
Convert sorted linked list to balanced BST (this can be done in place with O(m+n) time)
Here is what I would do.
This solution is O(n1+n2) time complexity.
STEPS:
Perform the inorder traversal of both the trees to get sorted arrays --> linear time
Merge the two arrays --> again linear time
Convert the merged array into a Balanced binary search tree --> again linear time
This would require O(n1+n2) time and space.
Links you may find useful while implementing:
How to merge 2 sorted arrays
Inorder traversal
Sorted array to a balanced BST
The following algorithm is from Algorithms in C++.
The idea is almost the same as in the algorithm posted by PengOne. This algorithm does in place merging, time complexity is O(n+m).
link join(link a, link b) {
if (b == 0) return a;
if (a == 0) return b;
insert(b, a->item);
b->left = join(a->left, b->left);
b->right = join(a->right, b->right);
delete a;
return b;
}
insert just inserts an item in the right place in the tree.
void insert(link &h, Item x) {
if (h == 0) {
h = new node(x);
return;
}
if (x.key() < h->item.key()) {
insert(h->left, x);
rotateRight(h);
}
else {
insert(h->right, x);
rotateLeft(h);
}
}
rotateRight and rotateLeft keep tree in the right order.
void rotateRight(link &h) {
link x = h->left;
h->left = x->right;
x->right = h;
h = x;
}
void rotateLeft(link &h) {
link x = h->right;
h->right = x->left;
x->left = h;
h = x;
}
Here link is node *.
Assuming the question is just to print sorted from both BSTs. Then the easier way is,
Store inorder traversal of 2 BSTs in 2 seperate arrays.
Now the problem reduces to merging\printing elements from 2 sorted arrays, which we got from step one. This merging can be done in o(m) when m>n or o(n) when m
Complexity: o(m+n)
Aux space: o(m+n) for the 2 arrays
MergeTwoBST_to_BalancedBST.java
public class MergeTwoBST_to_BalancedBST {
// arr1 and arr2 are the input arrays to be converted into a binary search
// structure and then merged and then balanced.
int[] arr1 = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 };
int[] arr2 = new int[] { 11, 12, 13, 14, 15, 16, 17, 18 };
BSTNode root1;
BSTNode root2;
// vector object to hold the nodes from the merged unbalanced binary search
// tree.
Vector<BSTNode> vNodes = new Vector<BSTNode>();
/**
* Constructor to initialize the Binary Search Tree root nodes to start
* processing. This constructor creates two trees from two given sorted
* array inputs. root1 tree from arr1 and root2 tree from arr2.
*
* Once we are done with creating the tree, we are traversing the tree in
* inorder format, to verify whether nodes are inserted properly or not. An
* inorder traversal should give us the nodes in a sorted order.
*/
public MergeTwoBST_to_BalancedBST() {
// passing 0 as the startIndex and arr1.length-1 as the endIndex.
root1 = getBSTFromSortedArray(arr1, 0, arr1.length - 1);
System.out.println("\nPrinting the first binary search tree");
inorder(root1); // traverse the tree in inorder format to verify whether
// nodes are inserted correctly or not.
// passing 0 as the startIndex and arr2.length-1 as the endIndex.
root2 = getBSTFromSortedArray(arr2, 0, arr2.length - 1);
System.out.println("\nPrinting the second binary search tree");
inorder(root2); // same here - checking whether the nodes are inserted
// properly or not.
}
/**
* Method to traverse the tree in inorder format. Where it traverses the
* left child first, then root and then right child.
*
* #param node
*/
public void inorder(BSTNode node) {
if (null != node) {
inorder(node.getLeft());
System.out.print(node.getData() + " ");
inorder(node.getRight());
}
}
/**
* Method to traverse the tree in preorder format. Where it traverses the
* root node first, then left child and then right child.
*
* #param node
*/
public void preorder(BSTNode node) {
if (null != node) {
System.out.print(node.getData() + " ");
preorder(node.getLeft());
preorder(node.getRight());
}
}
/**
* Creating a new Binary Search Tree object from a sorted array and
* returning the root of the newly created node for further processing.
*
* #param arr
* #param startIndex
* #param endIndex
* #return
*/
public BSTNode getBSTFromSortedArray(int[] arr, int startIndex, int endIndex) {
if (startIndex > endIndex) {
return null;
}
int middleIndex = startIndex + (endIndex - startIndex) / 2;
BSTNode node = new BSTNode(arr[middleIndex]);
node.setLeft(getBSTFromSortedArray(arr, startIndex, middleIndex - 1));
node.setRight(getBSTFromSortedArray(arr, middleIndex + 1, endIndex));
return node;
}
/**
* This method involves two operation. First - it adds the nodes from root1
* tree to root2 tree, and hence we get a merged root2 tree.Second - it
* balances the merged root2 tree with the help of a vector object which can
* contain objects only of BSTNode type.
*/
public void mergeTwoBinarySearchTree() {
// First operation - merging the trees. root1 with root2 merging should
// give us a new root2 tree.
addUtil(root1);
System.out.println("\nAfter the root1 tree nodes are added to root2");
System.out.println("Inorder Traversal of root2 nodes");
inorder(root2); // inorder traversal of the root2 tree should display
// the nodes in a sorted order.
System.out.println("\nPreorder traversal of root2 nodes");
preorder(root2);
// Second operation - this will take care of balancing the merged binary
// search trees.
balancedTheMergedBST();
}
/**
* Here we are doing two operations. First operation involves, adding nodes
* from root2 tree to the vector object. Second operation involves, creating
* the Balanced binary search tree from the vector objects.
*/
public void balancedTheMergedBST() {
// First operation : adding nodes to the vector object
addNodesToVector(root2, vNodes);
int vSize = vNodes.size();
// Second operation : getting a balanced binary search tree
BSTNode node = getBalancedBSTFromVector(vNodes, 0, vSize - 1);
System.out
.println("\n********************************************************");
System.out.println("After balancing the merged trees");
System.out.println("\nInorder Traversal of nodes");
inorder(node); // traversing the tree in inoder process should give us
// the output in sorted order ascending
System.out.println("\nPreorder traversal of root2 nodes");
preorder(node);
}
/**
* This method will provide us a Balanced Binary Search Tree. Elements of
* the root2 tree has been added to the vector object. It is parsed
* recursively to create a balanced tree.
*
* #param vNodes
* #param startIndex
* #param endIndex
* #return
*/
public BSTNode getBalancedBSTFromVector(Vector<BSTNode> vNodes,
int startIndex, int endIndex) {
if (startIndex > endIndex) {
return null;
}
int middleIndex = startIndex + (endIndex - startIndex) / 2;
BSTNode node = vNodes.get(middleIndex);
node.setLeft(getBalancedBSTFromVector(vNodes, startIndex,
middleIndex - 1));
node.setRight(getBalancedBSTFromVector(vNodes, middleIndex + 1,
endIndex));
return node;
}
/**
* This method traverse the tree in inorder process and adds each node from
* root2 to the vector object vNodes object only accepts objects of BSTNode
* type.
*
* #param node
* #param vNodes
*/
public void addNodesToVector(BSTNode node, Vector<BSTNode> vNodes) {
if (null != node) {
addNodesToVector(node.getLeft(), vNodes);
// here we are adding the node in the vector object.
vNodes.add(node);
addNodesToVector(node.getRight(), vNodes);
}
}
/**
* This method traverse the root1 tree in inorder process and add the nodes
* in the root2 tree based on their value
*
* #param node
*/
public void addUtil(BSTNode node) {
if (null != node) {
addUtil(node.getLeft());
mergeToSecondTree(root2, node.getData());
addUtil(node.getRight());
}
}
/**
* This method adds the nodes found from root1 tree as part it's inorder
* traversal and add it to the second tree.
*
* This method follows simple Binary Search Tree inserstion logic to insert
* a node considering the tree already exists.
*
* #param node
* #param data
* #return
*/
public BSTNode mergeToSecondTree(BSTNode node, int data) {
if (null == node) {
node = new BSTNode(data);
} else {
if (data < node.getData()) {
node.setLeft(mergeToSecondTree(node.getLeft(), data));
} else if (data > node.getData()) {
node.setRight(mergeToSecondTree(node.getRight(), data));
}
}
return node;
}
/**
*
* #param args
*/
public static void main(String[] args) {
MergeTwoBST_to_BalancedBST mergeTwoBST = new MergeTwoBST_to_BalancedBST();
mergeTwoBST.mergeTwoBinarySearchTree();
}
}
BSTNode.java:
public class BSTNode {
BSTNode left, right;
int data;
/* Default constructor */
public BSTNode() {
left = null;
right = null;
data = 0;
}
/* Constructor */
public BSTNode(int data) {
left = null;
right = null;
this.data = data;
}
public BSTNode getLeft() {
return left;
}
public void setLeft(BSTNode left) {
this.left = left;
}
public BSTNode getRight() {
return right;
}
public void setRight(BSTNode right) {
this.right = right;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
}

Reversing a singly linked list when a block size is given

There is a singly connected linked list and a block size is given.For eg if my linked list is 1->2->3->4->5->6->7->8-NULL and my block size is 4 then reverse the first 4 elements and then the second 4 elements.The output of the problem should be 4->3->2->1->8->7->6->5-NULL
I was thinking of dividing the linked list into segments of size 4 and then reversing it.
But that way I am forced to use a lot of extra nodes which is not desired at all.
The space complexity should be kept to a minimum.
It will be highly appreciable if someone can come with a better solution where the usage of extra nodes would be kept to a minimum.
I tried this...seems to work fine...
node* reverse(node* head) // function to reverse a list
{
node* new_head = NULL;
while(head != NULL)
{
node* next = head->next;
head->next = new_head;
new_head = head;
head = next;
}
return new_head;
}
node* reverse_by_block(node* head, int block)
{
if(head == NULL)
return head;
node* tmp = head;
node* new_head = head;
node* new_tail = NULL;
int count = block;
while(tmp != NULL && count--)
{
new_tail = tmp;
tmp = tmp->next;
}
new_tail->next = NULL;
new_tail = new_head;
new_head = reverse(new_head);
new_tail->next = reverse_by_block(tmp,block);
return new_head;
}
You can advance swapping the current element with the next 3 times: 1234, 2134, 2314, 2341. Then do it twice to get 3421. Then once to get 4321. Then advance 4 steps and repeat the process with the next block.
This can be done in linear-time, with constant space.
Here is a brief description:
Split the linked list into two parts by block-size
int split(node* head, node** listA, node** listB size_t block_size)
{
node* cur = head;
while(block_size && cur)
{
cur = cur->next;
--block_size;
}
if(!cur) { /* error : invalid block size */ return -1; }
*listA = head;
*listB = cur->next;
cur->next = NULL; /* terminate list A */
return 0;
}
Reverse the two sub-parts, (use a non-recursive linear time, constant space function)
reverse(listA);
reverse(listB);
Link them to get the desired linked list.
cur = *listA;
/* goto last but one element of listA */
while(cur->next) cur = cur->next;
cur->next = *listB;

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