Given a Binary Search Tree (BST) and a positive integer k, find the k’th smallest element in the Binary Search Tree.

For example, in the following BST, if k = 3, then output should be 10, and if k = 5, then output should be 14.

We have discussed two methods in this post and one method in this post. All of the previous methods require extra space. How to find the k’th largest element without extra space?

The idea is to use Morris Traversal. In this traversal, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree. See this for more details.

Below is C++ implementation of the idea.

// C++ program to find k'th largest element in BST #include<iostream> #include<climits> using namespace std; // A BST node struct Node { int key; Node *left, *right; }; // A function to find int KSmallestUsingMorris(Node *root, int k) { // Count to iterate over elements till we // get the kth smallest number int count = 0; int ksmall = INT_MIN; // store the Kth smallest Node *curr = root; // to store the current node while (curr != NULL) { // Like Morris traversal if current does // not have left child rather than printing // as we did in inorder, we will just // increment the count as the number will // be in an increasing order if (curr->left == NULL) { count++; // if count is equal to K then we found the // kth smallest, so store it in ksmall if (count==k) ksmall = curr->key; // go to current's right child curr = curr->right; } else { // we create links to Inorder Successor and // count using these links Node *pre = curr->left; while (pre->right != NULL && pre->right != curr) pre = pre->right; // building links if (pre->right==NULL) { //link made to Inorder Successor pre->right = curr; curr = curr->left; } // While breaking the links in so made temporary // threaded tree we will check for the K smallest // condition else { // Revert the changes made in if part (break link // from the Inorder Successor) pre->right = NULL; count++; // If count is equal to K then we found // the kth smallest and so store it in ksmall if (count==k) ksmall = curr->key; curr = curr->right; } } } return ksmall; //return the found value } // A utility function to create a new BST node Node *newNode(int item) { Node *temp = new Node; temp->key = item; temp->left = temp->right = NULL; return temp; } /* A utility function to insert a new node with given key in BST */ Node* insert(Node* node, int key) { /* If the tree is empty, return a new node */ if (node == NULL) return newNode(key); /* Otherwise, recur down the tree */ if (key < node->key) node->left = insert(node->left, key); else if (key > node->key) node->right = insert(node->right, key); /* return the (unchanged) node pointer */ return node; } // Driver Program to test above functions int main() { /* Let us create following BST 50 / \ 30 70 / \ / \ 20 40 60 80 */ Node *root = NULL; root = insert(root, 50); insert(root, 30); insert(root, 20); insert(root, 40); insert(root, 70); insert(root, 60); insert(root, 80); for (int k=1; k<=7; k++) cout << KSmallestUsingMorris(root, k) << " "; return 0; }

Output:

20 30 40 50 60 70 80

This article is contributed by Abhishek Somani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above