I'm still getting used to data structures, and I'm comfortable with traversing binary trees in the various ways, but I'm presented now with a situation where I have a normal binary tree, constructed of nodes that only know have data, left and right attributes.
However I want to transfer it into a "smarter" binary tree. This tree is to know its parent node, its total subnodes, and the level in the total tree it is at.
I'm really struggling with how I'd go about transferring the one "dumber" tree into the smarter version. My first instinct is to traverse recursively, but I'm not sure how I'd then be able to distinguish the parent and the level.
Copy the old tree to a new tree, using the normal recursive methods to traverse the original.
Since you're adding new attributes to the nodes, I presume you'll need to construct new nodes with fields for the new attributes.
Define a recursive function to copy the (sub)tree rooted at a given node. It needs as input its depth and parent. (The parent, of course, needs to be what will be its parent in the new tree.) Let it return the root of the new (sub)tree.
function copy_node (old_node, new_parent, depth) -> returns new_node {
new_node = new node
new_node.data = old_root.data // whatever that data might be
new_node.depth = depth
new_node.parent = parent
new_node.left = copy_node (old_node.left, new_node, depth + 1)
new_node.right = copy_node (old_node.right, new_node, depth + 1)
return new_node }
Copy the whole tree with
new_tree = copy_node (old_tree, nil, 0)
If you're using a language where fields can be added to existing objects willy-nilly, you don't even have to do the extra copying:
function adorn_node (node, parent, depth) {
node.parent = parent
node.depth = depth
adorn_node (node.left, node, depth + 1)
adorn_node (node.right, node, depth + 1) }
and start the ball rolling with
adorn_node (root, nil, 0)
That having been said, you will probably discover that there is a very good reason why most binary tree implementations do not contain these extra fields. It's a lot of work to maintain them across the many different operations you want to perform on trees. depth, especially, is hard to keep correct when you need to re-balance a tree.
And the fields don't generally buy you anything. Most algorithms that operate on trees do so using recursive functions, and as you can see from the above examples it's really easy to re-calculate both parent and depth on the fly while you're walking the tree. They don't need to be stored in the nodes themselves.
Tree-balancing often needs to know the difference in heights of the left and right subtrees. ("depth" is the distance to the root; "height" is the distance to the most distant leaf node in the subtree.) height is not so easy to calculate on the way down from the root, but fortunately you're usually only interested in which of the subtrees has the greatest height, and for that it's usually sufficient to store only the values -1, 0, +1 in each node.
I faced an interesting problem called as the Great Tree-List Problem. The Problem is as follows :
In the ordered binary tree, each node contains a single data element and "small" and "large" pointers to sub-trees .All the nodes in the "small" sub-tree are less than or equal to the data in the parent node. All the nodes in the "large" sub-tree are greater than the parent node. And a circular doubly linked list consist of previous and next pointers.
The problem is to take an ordered binary tree and rearrange the internal pointers to make a circular doubly linked list out of it. The "small" pointer should play the role of "previous" and the "large" pointer should play the role of "next". The list should be arranged so that the nodes are in increasing order. I have to write a recursive function & Return the head pointer to the new list.
The operation should be done in O(n) time.
I understand that recursion will go down the tree, but how to recursively change the small and large sub-trees into lists, also i have to append those lists together with the parent node.
How should i approach the problem?.. I just need a direction to solve the problem!.
The idea is to create a method that converts a tree node containing subtrees (children nodes) into a loop. And given a node that has converted children (e.g. after recursive calls came back), you create a new loop by pointing the large pointer (next) of the largest node to the smallest node, and the small pointer of the smallest node to the largest node.
May not be complete, but it will be close to this:
tree_node {
small
large
}
convert(node){
//base case 1, if leaf node
if node.small == null && node.large == null
return (self, self)
//recursively convert children
if node.small !=null
smallest, larger = convert(node.small)
else
smallest = larger = self
if node.large !=null
smaller, largest = convert(node.large)
else
smaller = largest = self
//wrap the ends of the chain
largest.large = smallest
smallest.small = largest
//wrap the mid part
smaller.small = larger
larger.large = small
//return pointers to the absolute smallest and largest of this subtree
return (smallest, largest)
}
//actually doing it
convert(tree.root)
The key to recursive programming is to imagine you already have the solution.
So, you already have a function recLink(Tree t) which receives a pointer to a tree, turns that tree into a doubly-linked circular list and returns a pointer to the list's head (leftmost) node:
recLink( n={Node: left, elt, right}) = // pattern match tree to a full node
rt := recLink( right); // already
lt := recLink( left); // have it
n.right := rt; n.left := lt.left; // middle node
lt.left.right := n; rt.left.right := lt; // right edges
lt.left := rt.left; rt.left := n;
return lt;
Finish up with the edge cases (empty child branches etc.). :)
assuming you have a simple tree of 3 nodes
B <--- A ---> C
walk down the left and right sides, get the pointers for each node, then have
B -> C
B <- C
Since your tree is binary, it will be composed of 3 node "subtrees" that can recursively use this strategy.
I'm trying to write a function to remove a node from a binary tree. I haven't coded the function yet, and I am trying to think about the different conditions I should consider for removing a node. I am guessing that the possible conditions are:
The node has no children
The node has one child
The node has 2 children
In each of these cases what would be the algorithm to perform a delete function?
This is something you would find in any standard textbook about algorithms, but let's suppose you are interested in the unbalanced case (balanced trees usually performs some rebalancing operations called "rotations" after a removal) and you use the "obvious" datastructure (a tree_node structure that holds the value and two pointers to other tree_node):
No children: release the memory hold by the node and set the parent's child link that pointed to it as NULL;
One child: release the memory hold by the node and set the parent's child link that pointed to it as the address of its unique child;
Two children: this is indeed the "complicated" case. Find the rightmost node of the left child (or the leftmost node of the right child), take its value, remove it (it is "case 1", so it is easy and can be done recursively) and set the current node's value as the one of that node. This is O(tree_height) = O(n), but it is not a problem (at least in theory) because this would be neverthless the complexity of finding a node.
Does your tree have any additional properties?
Is it an AVL?
If not, there are some pretty obvious and straightforward ways to do what you want (which will depend on your data representation, as Vitalij said).
And if it is an AVL for example, there ALSO are some well known method for doing that (wikipedia will tell you more on that topic)
First task is to find whether node exists which will be done during search and rest of your conditions are correct.
Leaf node: set the parent's child (right/left) to NULL.
Has one child: Just set the child of the node to be deleted to its parent's child.
Has two children: Basically have to re-order the whole subtree here by pruning the subtree to by finding new children for the node to be deleted.
Assuming you are dealing with general binary trees, do the following,
Node has no child- ie it is a leaf : Conveniently delete it..
Node has one child - Make the parent of the node to be deleted parent of its child , then delete the node. ie, if A->Parent = B; C->Parent = A; and A has to be deleted, then 1. Make C->Parent = B; 2. Delete A;
Tricky one.... Yes, replacing the node to be deleted by the left most child of the right subtree work, or by the rightmost tree of the left subtree, either will do... because it can be seen like this,
When a node is deleted, it has to be replaced by a node which satisfies some properties...
Lets say if our binary tree represents sorted numbers (in increasing order) in inorder traversal, then the deleted node should be replaced by some node from either of its subtrees. That should be larger in value than the whole remaining left subtree, and smaller than the whole remaining right subtree (remaining means the subtree remaining after adjusting for the deleted node successfully). Only two such nodes exist, leftmost leaf of the right subtree, or the rightmost node of left one.
Hence, replacing the deleted node from either one suffices...!!
Delete the given keys one at a time from the binary search tree. Possible equal keys were inserted into the left branch of the existing node. Please note that the insertion strategy also affects how the deletion is performed
BinarySearchTree-Delete
Node Delete(Node root, Key k)
1 if (root == null) // failed search
2 return null;
3 if (k == root.key) // successful search
4 return DeleteThis(root);
5 if (k < root.key) // k in the left branch
6 root.left = Delete(root.left, k);
7 else // k > root.key, i.e., k in the right branch
8 root.right = Delete(root.right, k);
9 return root;
Node DeleteThis(Node root)
1 if root has two children
2 p = Largest(root.left); // replace root with its immediate predecessor p
3 root.key = p.key;
4 root.left = Delete(root.left, p)
5 return root;
6 if root has only left child
7 return root.left
8 if root has only right child
9 return root.right
10 else root has no children
11 return null
Node Largest(Node root)
1 if root has no right child
2 return root
3 return Largest(root.right)
I am trying to solve a problem of finding a max repetitive sub-tree in an object tree.
By the object tree I mean a tree where each leaf and node has a name. Each leaf has a type and a value of that type associated with that leaf. Each node has a set of leaves / nodes in certain order.
Given an object tree that - we know - has a repetitive sub-tree in it.
By repetitive I mean 2 or more sub-trees that are similar in everything (names/types/order of sub-elements) but the values of leaves. No nodes/leaves can be shared between sub-trees.
Problem is to identify these sub-trees of the max height.
I know that the exhaustive search can do the trick. I am rather looking for more efficient approach.
you could implement a dfs traversal generating a hash value for each node. Store these values with the node height in a simple array. Sub-tree candidates are duplicate values, just check that the candidates are ok since two different sub-trees could yield same hash value.
Assuming the leafs and internal nodes are all of type Node and that standard access and traversal functions are available :
procedure dfs_update( node : Node, hashmap : Hashmap )
begin
if is_leaf(node) then
hashstring = concat("LEAF",'|',get_name_str(node),'|',get_type_str(node))
else // node is an internal node
hashstring = concat("NODE",'|',get_name_str(node))
for each child in get_children_sorted(node)
dfs_update(child,hashmap)
hashstring = concat(hashstring,'|',get_hash_string(hashmap,child))
end for
end if
// only a ref to node is added to the hashmap, we could also add
// the node's height, hashstring, whatever could be useful and inapropriate
// to keep in the Node ds
add(hashmap, hash(hashstring),node)
end
The tricky part is after a dfs_update, we have to get the list of collinding nodes in the hasmap by descending height and check two by two they are really repetitive.
What's the best way to create a balanced binary search tree from a sorted singly linked list?
How about creating nodes bottom-up?
This solution's time complexity is O(N). Detailed explanation in my blog post:
http://www.leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html
Two traversal of the linked list is all we need. First traversal to get the length of the list (which is then passed in as the parameter n into the function), then create nodes by the list's order.
BinaryTree* sortedListToBST(ListNode *& list, int start, int end) {
if (start > end) return NULL;
// same as (start+end)/2, avoids overflow
int mid = start + (end - start) / 2;
BinaryTree *leftChild = sortedListToBST(list, start, mid-1);
BinaryTree *parent = new BinaryTree(list->data);
parent->left = leftChild;
list = list->next;
parent->right = sortedListToBST(list, mid+1, end);
return parent;
}
BinaryTree* sortedListToBST(ListNode *head, int n) {
return sortedListToBST(head, 0, n-1);
}
You can't do better than linear time, since you have to at least read all the elements of the list, so you might as well copy the list into an array (linear time) and then construct the tree efficiently in the usual way, i.e. if you had the list [9,12,18,23,24,51,84], then you'd start by making 23 the root, with children 12 and 51, then 9 and 18 become children of 12, and 24 and 84 become children of 51. Overall, should be O(n) if you do it right.
The actual algorithm, for what it's worth, is "take the middle element of the list as the root, and recursively build BSTs for the sub-lists to the left and right of the middle element and attach them below the root".
Best isn't only about asynmptopic run time. The sorted linked list has all the information needed to create the binary tree directly, and I think this is probably what they are looking for
Note that the first and third entries become children of the second, then the fourth node has chidren of the second and sixth (which has children the fifth and seventh) and so on...
in psuedo code
read three elements, make a node from them, mark as level 1, push on stack
loop
read three elemeents and make a node of them
mark as level 1
push on stack
loop while top two enties on stack have same level (n)
make node of top two entries, mark as level n + 1, push on stack
while elements remain in list
(with a bit of adjustment for when there's less than three elements left or an unbalanced tree at any point)
EDIT:
At any point, there is a left node of height N on the stack. Next step is to read one element, then read and construct another node of height N on the stack. To construct a node of height N, make and push a node of height N -1 on the stack, then read an element, make another node of height N-1 on the stack -- which is a recursive call.
Actually, this means the algorithm (even as modified) won't produce a balanced tree. If there are 2N+1 nodes, it will produce a tree with 2N-1 values on the left, and 1 on the right.
So I think #sgolodetz's answer is better, unless I can think of a way of rebalancing the tree as it's built.
Trick question!
The best way is to use the STL, and advantage yourself of the fact that the sorted associative container ADT, of which set is an implementation, demands insertion of sorted ranges have amortized linear time. Any passable set of core data structures for any language should offer a similar guarantee. For a real answer, see the quite clever solutions others have provided.
What's that? I should offer something useful?
Hum...
How about this?
The smallest possible meaningful tree in a balanced binary tree is 3 nodes.
A parent, and two children. The very first instance of such a tree is the first three elements. Child-parent-Child. Let's now imagine this as a single node. Okay, well, we no longer have a tree. But we know that the shape we want is Child-parent-Child.
Done for a moment with our imaginings, we want to keep a pointer to the parent in that initial triumvirate. But it's singly linked!
We'll want to have four pointers, which I'll call A, B, C, and D. So, we move A to 1, set B equal to A and advance it one. Set C equal to B, and advance it two. The node under B already points to its right-child-to-be. We build our initial tree. We leave B at the parent of Tree one. C is sitting at the node that will have our two minimal trees as children. Set A equal to C, and advance it one. Set D equal to A, and advance it one. We can now build our next minimal tree. D points to the root of that tree, B points to the root of the other, and C points to the... the new root from which we will hang our two minimal trees.
How about some pictures?
[A][B][-][C]
With our image of a minimal tree as a node...
[B = Tree][C][A][D][-]
And then
[Tree A][C][Tree B]
Except we have a problem. The node two after D is our next root.
[B = Tree A][C][A][D][-][Roooooot?!]
It would be a lot easier on us if we could simply maintain a pointer to it instead of to it and C. Turns out, since we know it will point to C, we can go ahead and start constructing the node in the binary tree that will hold it, and as part of this we can enter C into it as a left-node. How can we do this elegantly?
Set the pointer of the Node under C to the node Under B.
It's cheating in every sense of the word, but by using this trick, we free up B.
Alternatively, you can be sane, and actually start building out the node structure. After all, you really can't reuse the nodes from the SLL, they're probably POD structs.
So now...
[TreeA]<-[C][A][D][-][B]
[TreeA]<-[C]->[TreeB][B]
And... Wait a sec. We can use this same trick to free up C, if we just let ourselves think of it as a single node instead of a tree. Because after all, it really is just a single node.
[TreeC]<-[B][A][D][-][C]
We can further generalize our tricks.
[TreeC]<-[B][TreeD]<-[C][-]<-[D][-][A]
[TreeC]<-[B][TreeD]<-[C]->[TreeE][A]
[TreeC]<-[B]->[TreeF][A]
[TreeG]<-[A][B][C][-][D]
[TreeG]<-[A][-]<-[C][-][D]
[TreeG]<-[A][TreeH]<-[D][B][C][-]
[TreeG]<-[A][TreeH]<-[D][-]<-[C][-][B]
[TreeG]<-[A][TreeJ]<-[B][-]<-[C][-][D]
[TreeG]<-[A][TreeJ]<-[B][TreeK]<-[D][-]<-[C][-]
[TreeG]<-[A][TreeJ]<-[B][TreeK]<-[D][-]<-[C][-]
We are missing a critical step!
[TreeG]<-[A]->([TreeJ]<-[B]->([TreeK]<-[D][-]<-[C][-]))
Becomes :
[TreeG]<-[A]->[TreeL->([TreeK]<-[D][-]<-[C][-])][B]
[TreeG]<-[A]->[TreeL->([TreeK]<-[D]->[TreeM])][B]
[TreeG]<-[A]->[TreeL->[TreeN]][B]
[TreeG]<-[A]->[TreeO][B]
[TreeP]<-[B]
Obviously, the algorithm can be cleaned up considerably, but I thought it would be interesting to demonstrate how one can optimize as you go by iteratively designing your algorithm. I think this kind of process is what a good employer should be looking for more than anything.
The trick, basically, is that each time we reach the next midpoint, which we know is a parent-to-be, we know that its left subtree is already finished. The other trick is that we are done with a node once it has two children and something pointing to it, even if all of the sub-trees aren't finished. Using this, we can get what I am pretty sure is a linear time solution, as each element is touched only 4 times at most. The problem is that this relies on being given a list that will form a truly balanced binary search tree. There are, in other words, some hidden constraints that may make this solution either much harder to apply, or impossible. For example, if you have an odd number of elements, or if there are a lot of non-unique values, this starts to produce a fairly silly tree.
Considerations:
Render the element unique.
Insert a dummy element at the end if the number of nodes is odd.
Sing longingly for a more naive implementation.
Use a deque to keep the roots of completed subtrees and the midpoints in, instead of mucking around with my second trick.
This is a python implementation:
def sll_to_bbst(sll, start, end):
"""Build a balanced binary search tree from sorted linked list.
This assumes that you have a class BinarySearchTree, with properties
'l_child' and 'r_child'.
Params:
sll: sorted linked list, any data structure with 'popleft()' method,
which removes and returns the leftmost element of the list. The
easiest thing to do is to use 'collections.deque' for the sorted
list.
start: int, start index, on initial call set to 0
end: int, on initial call should be set to len(sll)
Returns:
A balanced instance of BinarySearchTree
This is a python implementation of solution found here:
http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html
"""
if start >= end:
return None
middle = (start + end) // 2
l_child = sll_to_bbst(sll, start, middle)
root = BinarySearchTree(sll.popleft())
root.l_child = l_child
root.r_child = sll_to_bbst(sll, middle+1, end)
return root
Instead of the sorted linked list i was asked on a sorted array (doesn't matter though logically, but yes run-time varies) to create a BST of minimal height, following is the code i could get out:
typedef struct Node{
struct Node *left;
int info;
struct Node *right;
}Node_t;
Node_t* Bin(int low, int high) {
Node_t* node = NULL;
int mid = 0;
if(low <= high) {
mid = (low+high)/2;
node = CreateNode(a[mid]);
printf("DEBUG: creating node for %d\n", a[mid]);
if(node->left == NULL) {
node->left = Bin(low, mid-1);
}
if(node->right == NULL) {
node->right = Bin(mid+1, high);
}
return node;
}//if(low <=high)
else {
return NULL;
}
}//Bin(low,high)
Node_t* CreateNode(int info) {
Node_t* node = malloc(sizeof(Node_t));
memset(node, 0, sizeof(Node_t));
node->info = info;
node->left = NULL;
node->right = NULL;
return node;
}//CreateNode(info)
// call function for an array example: 6 7 8 9 10 11 12, it gets you desired
// result
Bin(0,6);
HTH Somebody..
This is the pseudo recursive algorithm that I will suggest.
createTree(treenode *root, linknode *start, linknode *end)
{
if(start == end or start = end->next)
{
return;
}
ptrsingle=start;
ptrdouble=start;
while(ptrdouble != end and ptrdouble->next !=end)
{
ptrsignle=ptrsingle->next;
ptrdouble=ptrdouble->next->next;
}
//ptrsignle will now be at the middle element.
treenode cur_node=Allocatememory;
cur_node->data = ptrsingle->data;
if(root = null)
{
root = cur_node;
}
else
{
if(cur_node->data (less than) root->data)
root->left=cur_node
else
root->right=cur_node
}
createTree(cur_node, start, ptrSingle);
createTree(cur_node, ptrSingle, End);
}
Root = null;
The inital call will be createtree(Root, list, null);
We are doing the recursive building of the tree, but without using the intermediate array.
To get to the middle element every time we are advancing two pointers, one by one element, other by two elements. By the time the second pointer is at the end, the first pointer will be at the middle.
The running time will be o(nlogn). The extra space will be o(logn). Not an efficient solution for a real situation where you can have R-B tree which guarantees nlogn insertion. But good enough for interview.
Similar to #Stuart Golodetz and #Jake Kurzer the important thing is that the list is already sorted.
In #Stuart's answer, the array he presented is the backing data structure for the BST. The find operation for example would just need to perform index array calculations to traverse the tree. Growing the array and removing elements would be the trickier part, so I'd prefer a vector or other constant time lookup data structure.
#Jake's answer also uses this fact but unfortunately requires you to traverse the list to find each time to do a get(index) operation. But requires no additional memory usage.
Unless it was specifically mentioned by the interviewer that they wanted an object structure representation of the tree, I would use #Stuart's answer.
In a question like this you'd be given extra points for discussing the tradeoffs and all the options that you have.
Hope the detailed explanation on this post helps:
http://preparefortechinterview.blogspot.com/2013/10/planting-trees_1.html
A slightly improved implementation from #1337c0d3r in my blog.
// create a balanced BST using #len elements starting from #head & move #head forward by #len
TreeNode *sortedListToBSTHelper(ListNode *&head, int len) {
if (0 == len) return NULL;
auto left = sortedListToBSTHelper(head, len / 2);
auto root = new TreeNode(head->val);
root->left = left;
head = head->next;
root->right = sortedListToBSTHelper(head, (len - 1) / 2);
return root;
}
TreeNode *sortedListToBST(ListNode *head) {
int n = length(head);
return sortedListToBSTHelper(head, n);
}
If you know how many nodes are in the linked list, you can do it like this:
// Gives path to subtree being built. If branch[N] is false, branch
// less from the node at depth N, if true branch greater.
bool branch[max depth];
// If rem[N] is true, then for the current subtree at depth N, it's
// greater subtree has one more node than it's less subtree.
bool rem[max depth];
// Depth of root node of current subtree.
unsigned depth = 0;
// Number of nodes in current subtree.
unsigned num_sub = Number of nodes in linked list;
// The algorithm relies on a stack of nodes whose less subtree has
// been built, but whose right subtree has not yet been built. The
// stack is implemented as linked list. The nodes are linked
// together by having the "greater" handle of a node set to the
// next node in the list. "less_parent" is the handle of the first
// node in the list.
Node *less_parent = nullptr;
// h is root of current subtree, child is one of its children.
Node *h, *child;
Node *p = head of the sorted linked list of nodes;
LOOP // loop unconditionally
LOOP WHILE (num_sub > 2)
// Subtract one for root of subtree.
num_sub = num_sub - 1;
rem[depth] = !!(num_sub & 1); // true if num_sub is an odd number
branch[depth] = false;
depth = depth + 1;
num_sub = num_sub / 2;
END LOOP
IF (num_sub == 2)
// Build a subtree with two nodes, slanting to greater.
// I arbitrarily chose to always have the extra node in the
// greater subtree when there is an odd number of nodes to
// split between the two subtrees.
h = p;
p = the node after p in the linked list;
child = p;
p = the node after p in the linked list;
make h and p into a two-element AVL tree;
ELSE // num_sub == 1
// Build a subtree with one node.
h = p;
p = the next node in the linked list;
make h into a leaf node;
END IF
LOOP WHILE (depth > 0)
depth = depth - 1;
IF (not branch[depth])
// We've completed a less subtree, exit while loop.
EXIT LOOP;
END IF
// We've completed a greater subtree, so attach it to
// its parent (that is less than it). We pop the parent
// off the stack of less parents.
child = h;
h = less_parent;
less_parent = h->greater_child;
h->greater_child = child;
num_sub = 2 * (num_sub - rem[depth]) + rem[depth] + 1;
IF (num_sub & (num_sub - 1))
// num_sub is not a power of 2
h->balance_factor = 0;
ELSE
// num_sub is a power of 2
h->balance_factor = 1;
END IF
END LOOP
IF (num_sub == number of node in original linked list)
// We've completed the full tree, exit outer unconditional loop
EXIT LOOP;
END IF
// The subtree we've completed is the less subtree of the
// next node in the sequence.
child = h;
h = p;
p = the next node in the linked list;
h->less_child = child;
// Put h onto the stack of less parents.
h->greater_child = less_parent;
less_parent = h;
// Proceed to creating greater than subtree of h.
branch[depth] = true;
num_sub = num_sub + rem[depth];
depth = depth + 1;
END LOOP
// h now points to the root of the completed AVL tree.
For an encoding of this in C++, see the build member function (currently at line 361) in https://github.com/wkaras/C-plus-plus-intrusive-container-templates/blob/master/avl_tree.h . It's actually more general, a template using any forward iterator rather than specifically a linked list.