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?

难度:medium

题目:给定二叉搜索树,找出其值第K小的结点。

思路:中序遍历

Runtime: 0 ms, faster than 100.00% of Java online submissions for Kth Smallest Element in a BST.
Memory Usage: 38.9 MB, less than 19.71% of Java online submissions for Kth Smallest Element in a BST.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int[] result = {root.val};
        kthSmallest(root, k, new int[1], result);
        
        return result[0];
    }
    
    public void kthSmallest(TreeNode root, int k, int[] count, int[] result) {
        if (root == null || count[0] >= k) {
            return;
        }
        kthSmallest(root.left, k, count, result);
        count[0]++;
        if (count[0] == k) {
            result[0] = root.val;
            return;
        } 
        kthSmallest(root.right, k, count, result);
    }
}

linm
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