# 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
Follow up: 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.

## 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);
}
}