I was asked this question, which I personally find hard:
Create a data structure that can:
Insert elements,
Remove elements,
Search Elements,
In time O(log n)
In addition,
It should have the following two functions which work in time O(1):
next(x):
given a pointer to the node of value x, return a pointer to the node with the smallest bigger value than x.
previous(x)
given a pointer to the node of value x, return a pointer to the node with the biggest smallest value than x.
If each node contains a pointer to its successor and a pointer to its predecessor, or equivalently - if you maintain both a doublely linked list and a tree, where each node in the tree points to its equivalent node in the list and vice versa - you'll get want you want. Updating the list on insert/delete is O(1) (after locating the closest node in the tree). Searching is performed on the tree. Succesor / predecessor are performed on the list.
#RogerLindsjö's idea from the comments is a good one. Essentially, keep a regular, balanced BST, then thread a doubly-linked list through the nodes keeping them in sorted order. That way, given a pointer to a node in the tree, you can find the largest value smaller than it or the smallest value greater than it simply by following the next or previous pointers in each node.
You can maintain this list through insertions and deletions without changing the overall runtime of an insert or delete. For example, here's how you might do an insertion of an element x:
Do a standard BST successor query to find the smallest value larger than x in the tree, and a standard BST predecessor query to find the largest value smaller than x in the tree. Each search takes time O(log n) to complete.
Do a regular BST insertion to insert x. Then, set its next and previous pointers to the two elements you found in the previous step, and update those nodes to point to your new node x. This also takes time O(log n).
The total time for the insertion is then O(log n), matching what a balanced tree can provide.
I'll leave it to you to figure out deletion, which can similarly maintain the linked list pointers without changing the overall cost of the operation.
Like most self-balancing trees, a B+ tree provides Insert, Remove, and Search operations with O(log n) time complexity.
In a B+ tree, a leaf node hosts multiple keys in an array, so the concept of "pointer to node with value x" does not really exist, but we could define it as the tuple (pointer, index), where the pointer is to the node, and index is the slot in which x is stored.
In a B+ tree the nodes at the bottom level contain all the keys, and these nodes are often linked, usually only in forward direction (i.e. to the right), but it is quite possible to also maintain a link in the opposite direction, without increasing the time complexity of the above operations.
With those two remarks in mind, prev-next operations can clearly be executed in O(1) time.
If your elements are integers you can use y-fast trie that supports all mentioned operations in O(log log m). Also, almost any search tree will allow doing these operations in O(log n) time by just going first up and then down (it will require a lot of concentration to not mess up with the order, though)
You can use two pointers in the node of the balanced tree, namely pred - predecessor and succ - successor. While inserting a node into the tree or deleting a node from the tree you just have to do some pointer manipulations, equivalent to those in doubly linked list.
The time complexity will be O(1) in each case.
I have provided my implementation for the insertion and deletion in case of AVL Tree below. The complete implementation is available here.
Structure of node
template<typename T>
struct node {
T key;
int freq;
node<T> *left;
node<T> *right;
node<T> *pred;
node<T> *succ;
int height;
node(T key): key(key), freq(1),
left(nullptr),
right(nullptr),
height(1),
pred(nullptr),
succ(nullptr) {}
};
insert function
node<T> *insert(node<T> *root, T key) {
if(root == nullptr)
return new node<T>(key);
if(!comp(key, root->key) && !comp(root->key, key)) {
++root->freq;
} else if(comp(key, root->key)) {
if(root->left == nullptr) {
node<T> *new_node = new node<T>(key);
/* set the pred and succ ptrs*/
new_node->succ = root;
new_node->pred = root->pred;
if(root->pred != nullptr)
root->pred->succ = new_node;
root->pred = new_node;
root->left = new_node;
} else {
root->left = insert(root->left, key);
}
} else {
if(root->right == nullptr) {
node<T> *new_node = new node<T>(key);
/* set the pred and succ ptrs*/
new_node->pred = root;
new_node->succ = root->succ;
if(root->succ != nullptr)
root->succ->pred = new_node;
root->succ = new_node;
root->right = new_node;
} else {
root->right = insert(root->right, key);
}
}
root->height = max(height(root->left), height(root->right)) + 1;
int bf = balance_factor(root);
node<T> *left = root->left;
node<T> *right = root->right;
if(bf > 1 && left != nullptr && comp(key, left->key)) {
/*
node was inserted at left subtree of left child
fix - right rotate root
*/
root = rotate_right(root);
} else if(bf > 1 && left != nullptr && comp(left->key, key)) {
/*
node was inserted at right subtree of left child
fix - left rotate left child
- right rotate root
*/
root->left = rotate_left(root->left);
root = rotate_right(root);
} else if(bf < -1 && right != nullptr && comp(right->key, key)) {
/*
node was inserted at right subtree of right child
fix - left rotate root
*/
root = rotate_left(root);
} else if(bf < -1 && right != nullptr && comp(key, right->key)) {
/*
node was inserted at left subtree of right child
fix - right rotate right child
- left roatate root
*/
root->right = rotate_right(root->right);
root = rotate_left(root);
}
return root;
}
erase function
node<T> *erase(node<T> *root, T key) {
if(root == nullptr)
return nullptr;
if(comp(key, root->key)) {
root->left = erase(root->left, key);
} else if(comp(root->key, key)) {
root->right = erase(root->right, key);
} else {
if(root->left == nullptr || root->right == nullptr) {
/* update pred and succ ptrs */
if(root->succ != nullptr)
root->succ->pred = root->pred;
if(root->pred != nullptr)
root->pred->succ = root->succ;
if(root->right == nullptr) {
node<T> *temp = root->left;
delete root;
root = temp;
} else {
node<T> *temp = root->right;
delete root;
root = temp;
}
} else {
// node<T> *succ = minVal(root->right);
root->key = root->succ->key;
root->freq = root->succ->freq;
root->right = erase(root->right, root->succ->key);
}
}
if(root != nullptr) {
root->height = max(height(root->left), height(root->right)) + 1;
int bf_root = balance_factor(root);
if(bf_root > 1) {
/*
case R
*/
int bf_left = balance_factor(root->left);
if(bf_left >= 0) {
/*
case R0 and R1
*/
root = rotate_right(root);
} else {
/*
case R -1
*/
root->left = rotate_left(root->left);
root = rotate_right(root);
}
} else if(bf_root < -1) {
/*
Case L
*/
int bf_right = balance_factor(root->right);
if(bf_right <= 0) {
/*
case L0 and L -1
*/
root = rotate_left(root);
} else {
/*
case L1
*/
root->right = rotate_right(root->right);
root = rotate_left(root);
}
}
}
return root;
}
I am trying to learn DSA and got stuck on one problem.
How to calculate height of a tree. I mean normal tree, not any specific implementation of tree like BT or BST.
I have tried google but seems everyone is talking about Binary tree and nothing is available for normal tree.
Can anyone help me to redirect to some page or articles to calculate height of a tree.
Lets say a typical node in your tree is represented as Java class.
class Node{
Entry entry;
ArrayList<Node> children;
Node(Entry entry, ArrayList<Node> children){
this.entry = entry;
this.children = children;
}
ArrayList<Node> getChildren(){
return children;
}
}
Then a simple Height Function can be -
int getHeight(Node node){
if(node == null){
return 0;
}else if(node.getChildren() == null){
return 1;
} else{
int childrenMaxHeight = 0;
for(Node n : node.getChildren()){
childrenMaxHeight = Math.max(childrenMaxHeight, getHeight(n));
}
return 1 + childrenMaxHeight;
}
}
Then you just need to call this function passing the root of tree as argument. Since it traverse all the node exactly once, the run time is O(n).
1. If height of leaf node is considered as 0 / Or height is measured depending on number of edges in longest path from root to leaf :
int maxHeight(treeNode<int>* root){
if(root == NULL)
return -1; // -1 beacuse since a leaf node is 0 then NULL node should be -1
int h=0;
for(int i=0;i<root->childNodes.size();i++){
temp+=maxHeight(root->childNodes[i]);
if(temp>h){
h=temp;
}
}
return h+1;
}
2. If height of root node is considered 1:
int maxHeight(treeNode<int>* root){
if(root == NULL)
return 0;
int h=0;
for(int i=0;i<root->childNodes.size();i++){
temp+=maxHeight(root->childNodes[i]);
if(temp>h){
h=temp;
}
}
return h+1;
Above Code is based upon following class :
template <typename T>
class treeNode{
public:
T data;
vector<treeNode<T>*> childNodes; // vector for storing pointer to child treenode
creating Tree node
treeNode(T data){
this->data = data;
}
};
In case of 'normal tree' you can recursively calculate the height of tree in similar fashion to a binary tree but here you will have to consider all children at a node instead of just two.
To find a tree height a BFS iteration will work fine.
Edited form Wikipedia:
Breadth-First-Search(Graph, root):
create empty set S
create empty queues Q1, Q2
root.parent = NIL
height = -1
Q1.enqueue(root)
while Q1 is not empty:
height = height + 1
switch Q1 and Q2
while Q2 is not empty:
for each node n that is adjacent to current:
if n is not in S:
add n to S
n.parent = current
Q1.enqueue(n)
You can see that adding another queue allows me to know what level of the tree.
It iterates for each level, and for each mode in that level.
This is a discursion way to do it (opposite of recursive). So you don't have to worry about that too.
Run time is O(|V|+ |E|).
Problem:
Let T be a binary tree. Define a Roman node to be a node v in T, such that the number of descendants in v’s left subtree differs from the number of nodes in v’s right subtree by at most 5. Describe a linear-time algorithm for finding each node v of T, such that v is not a Roman node, but all of v’s descendants are Roman nodes.
What I have so far:
I could think of O(n^2) (top-down approach) solution where I will traverse the tree and check if a node is not roman, then traverse this node's descendants to check whether all of them are roman or not. So, in this way I am traversing each node twice.
I am assuming there is a bottom-up approach where it is possible to find the required node in O(n).
Any ideas?
Thanks #n.m. for your comments. I have implemented a solution using your inputs.
Logic:
The basic logic is to set isRoman field of a node to false even when a node's descendant (suppose grand grand child) is not roman. This means that even if a node satisfies the property (left and right descendants are at most 5) but if its left and right subtree's isRoman is false, then current node's isRoman will also be false.
/*
* public class TreeNode {
* int val;
* int size; //stores the number of descendants including itself
* boolean isRoman; // is true when this node is roman and all its descendants are roman,
* //false even when a grand grand child node of this node is not roman.
* TreeNode left;
* TreeNode right;
* }
*/
public static void roman(TreeNode root, List<TreeNode> lst){
if(root == null)
return;
if(root.left == null && root.right == null){
root.size = 1;
root.isRoman = true;
return;
}
int left = 0;
int right = 0;
boolean isLeftRoman = false;
boolean isRightRoman = false;
if(root.left != null) {
roman(root.left,lst);
left = root.left.size;
isLeftRoman = root.left.isRoman;
}else{
isLeftRoman=true;
}
if(root.right != null) {
roman(root.right,lst);
right = root.right.size;
isRightRoman = root.right.isRoman;
}else{
isRightRoman=true;
}
//if the current node is roman and all it's descendants are roman, update isRoman to true
if(Math.abs(left-right) <= 5 && isLeftRoman && isRightRoman)
root.isRoman = true;
else
root.isRoman = false;
//update the current node's size
root.size = left+right+1;
//add the node to list which is not roman but all its descendants are roman
if(Math.abs(left-right) > 5 && isLeftRoman && isRightRoman)
lst.add(root);
}
Suppose i am having an array say
1 5 4 6 8 9 10 22 17 7 9 3
I want to create a binary search tree from this array. I need algorithm to understand that.
I have read rest other things related to BST like inorder traversal preorder postorder, tree walk, insertion deletion etc
Book has not provided how to create BST. Need help here
if you do not care about the tree being balanced it is simple:
put the first element of the tree as the head.
iterate over the array. if an element is bigger than the node take a left(repeat the step for the left child) otherwise take a right(repeat the step for the right child).
if the left/right child is a null insert your new value there.
guaranteed to produce a binary search tree - just not a balanced one.
Firstly, you should choose a root node for your BST. Once you have chosen a root node, it is already easy to construct a BST taking into consideration the fact that: left children are less than the parent node and all right children are greater than the parent node.
private Node root;
public void insert(int val) {
if (root == null) {
root = new Node(val);
} else {
insertHelper(root, val);
}
}
private void insertHelper(Node node, int val) {
if (val < node.val) {
if (node.left == null) {
node.left = new Node(val);
} else {
insertHelper(node.left, val);
}
} else if (node.val < val) {
if (node.right == null) {
node.right = new Node(val);
} else {
insertHelper(node.right, val);
}
}
}
If the given array is sorted, you can do the following:
Take the middle element of the array and make it the root of the tree
Take the left sub-array and make it the left sub-tree of the root recursively
Take the right sub-array and make it the right sub-tree of the root recursively
Otherwise you can always sort the array before applying the procedure
struct node* construct(int arr[], int start, int end)
{
if(start>end)
return;
else if (start==end)
{
/*assuming we have a function newNode(int) which creates a new BST Node*/
node* Node = newNode(arr[start]);
return Node;
}
int mid = (start+end)/2;
node* root = newNode(arr[mid]);
root->left = construct(arr,start,mid-1);
root->right = construct(arr,mid+1,end);
return root;
}
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;
}
}