## Invert given Binary Tree | Recursive and Iterative solution

Given a binary tree, write an efficient algorithm to invert binary tree.

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Given a binary tree, write an efficient algorithm to invert binary tree.

Convert a given binary tree to BST (Binary Search Tree) by keeping original structure of the binary tree intact.

Given a normal binary tree, convert it to Left-child right-sibling (LC-RS) binary tree.

Huffman Coding is a algorithm for doing data compression and it forms the basic idea behind file compression. This post talks about fixed length and variable length encoding, uniquely decodable codes, prefix rules and construction of Huffman Tree.

Given a binary tree, write an efficient algorithm to check if tree is height balanced or not. In a height balanced tree, the absolute difference between height of left subtree and right subtree for every node is 0 or 1.

Given a BST, find ceil and floor of a given key in it. If the given key lie in the BST, then both floor and ceil is equal to that key, else ceil is equal to next greater key (if any) in the BST and floor is equal to previous greater key (if any) in …

Given a Binary Search Tree and a positive number K, find K’th smallest and K’th largest element in BST.

Given a BST and two nodes x and y in it, find lowest common ancestor (LCA) of x and y in it.

Given a BST, find in-order predecessor of a given key in it. If the given key do not lie in the BST, then return the previous greater key (if any) present in the BST. An in-order predecessor of a node in BST is the previous node in in-order traversal of it.

Given two arrays which represents set of BST keys, check if they represents same BSTs or not. We are not allowed to build the tree.

Given a binary tree, calculate sum of all nodes for each diagonal having negative slope (\). Assume that the left and right child of a node makes 45 degree angle with the parent.

Given a Binary Tree, determine if it is a BST or not. This problem has a simple recursive solution. The BST property “every node on the right subtree has to be larger than the current node and every node on the left subtree has to be smaller than the current node” is the key …