# Kth Smallest Element in a BST

Given a binary search tree, write a function`kthSmallest`to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Example 1:

``````Input:
root = [3,1,4,null,2], k = 1
3
/ \
1   4
\
2

Output:
1
``````

Example 2:

``````Input:
root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3   6
/ \
2   4
/
1

Output:
3
``````

What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

## Analysis

In order traversal of BST actually returns the element in ascending order, thus intuitively, traverse the BST with in-order, and return the kth element in the result, would be the kth smallest element in a BST.

https://leetcode.com/problems/kth-smallest-element-in-a-bst/discuss/63660/3-ways-implemented-in-JAVA-\(Python\):-Binary-Search-in-order-iterative-and-recursive

## Solution

DFS in order traverse

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int kthSmallest(TreeNode root, int k) {
if (root == null) return 0;
List<Integer> topK = new ArrayList<Integer>();
helper(root, topK, k);
}
private void helper(TreeNode root, List<Integer> topK, int k) {
if (root == null) return;
helper(root.left, topK, k);
if (topK.size() < k) {
} else {
return;
}
helper(root.right, topK, k);
}
}
``````
``````class Solution {

int count = 0;
int result = 0;
public int kthSmallest(TreeNode root, int k) {
count = 0;
result = 0;
dfs(root, k);
return result;
}

boolean dfs(TreeNode x, int k) {
if (x == null) return false;

if (dfs(x.left, k)) {
return true;
}

count++;
if (count == k) {
result = x.val;
return true;
}

return dfs(x.right, k);
}
}
``````