Given the root of a binary search tree, and an integer k, return the kth (1-indexed) smallest element in the tree.
Example 1:
Input: root = [3,1,4,null,2], k = 1
Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3
Constraints:
n.1 <= k <= n <= 1040 <= Node.val <= 104Kth Smallest Element in a BST - LeetCode
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
// The basic idea is to perform an inorder traversal of the BST and keep a count variable (initilaized to 0)
// which when gets incremented when it encounted a root node in the traversal. If its value becomes
// equal to k return that node's value.
int count = 0;
void solve(TreeNode *root,int k,int &res)
{
if(!root)
return;
solve(root->left,k,res);
count++;
if(count==k)
{
res = root->val;
return;
}
solve(root->right,k,res);
}
int kthSmallest(TreeNode* root, int k) {
int res = -1;
solve(root,k,res);
return res;
}
};