Find k-th smallest element in BST (Order Statistics in BST) in C++

Suppose we have a binary search tree and a value K as input, we have to find K-th smallest element in the tree.

So, if the input is like

k = 3, then the output will be 15.

To solve this, we will follow these steps −

• Define a function find_kth_smallest(), this will take root, count, k,

• if root is NULL, then −

  • return NULL

• left = find_kth_smallest(left of root, count, k)

• if left is not NULL, then −

  • return left

• (increase count by 1)

• if count is same as k, then −

  • return root

• return find_kth_smallest(right of root, count, k)

• From the main method, do the following −

• count := 0

• res = find_kth_smallest(root, count, k)

• if res is NULL, then −

  • display not found

• Otherwise

  • display val of res

Example (C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <iostream>
using namespace std;
struct TreeNode {
   int val;
   TreeNode *left, *right;
   TreeNode(int x) {
      val = x;
      left = right = NULL;
   }
};
TreeNode* find_kth_smallest(TreeNode* root, int &count, int k) {
   if (root == NULL)
      return NULL;
   TreeNode* left = find_kth_smallest(root->left, count, k);
   if (left != NULL)
      return left;
   count++;
   if (count == k)
      return root;
   return find_kth_smallest(root->right, count, k);
}
void kth_smallest(TreeNode* root, int k) {
   int count = 0;
   TreeNode* res = find_kth_smallest(root, count, k);
   if (res == NULL)
      cout << "Not found";
   else
      cout << res->val;
}
int main() {
   TreeNode* root = new TreeNode(25);
   root->left = new TreeNode(13);
   root->right = new TreeNode(27);
   root->left->left = new TreeNode(9);
   root->left->right = new TreeNode(17);
   root->left->right->left = new TreeNode(15);
   root->left->right->right = new TreeNode(19);
   int k = 3;
   kth_smallest(root, k);
}

Input

TreeNode* root = new TreeNode(25); root->left = new TreeNode(13);
root->right = new TreeNode(27); root->left->left = new
TreeNode(9); root->left->right = new TreeNode(17); root- >left->right->left = new TreeNode(15); root->left->right->right = new TreeNode(19); k = 3

Output

15