## 题目地址

https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/

## 题目描述

```
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?

```

## 思路

解法一：

由于‘中序遍历一个二叉查找树（BST）的结果是一个有序数组’ ，因此我们只需要在遍历到第k个，返回当前元素即可。
中序遍历相关思路请查看[binary-tree-traversal](../thinkings/binary-tree-traversal.md)

解法二：

联想到二叉搜索树的性质，root 大于左子树，小于右子树，如果左子树的节点数目等于 K-1，那么 root 就是结果，否则如果左子树节点数目小于 K-1，那么结果必然在右子树，否则就在左子树。
因此在搜索的时候同时返回节点数目，跟 K 做对比，就能得出结果了。


## 关键点解析

- 中序遍历

## 代码

解法一：

JavaScript Code:

```js


/*
 * @lc app=leetcode id=230 lang=javascript
 *
 * [230] Kth Smallest Element in a BST
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function(root, k) {
    const stack = [root];
    let cur = root;
    let i = 0;

    function insertAllLefts(cur) {
        while(cur && cur.left) {
            const l = cur.left;
            stack.push(l);
            cur = l;
        }
    }
    insertAllLefts(cur);

    while(cur = stack.pop()) {
        i++;
        if (i === k) return cur.val;
        const r = cur.right;

        if (r) {
            stack.push(r);
            insertAllLefts(r);
        }
    }

    return -1;

    
};
```

Java Code:

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
private int count = 1;
private int res;

public int KthSmallest (TreeNode root, int k) {
    inorder(root, k);
    return res;
}

public void inorder (TreeNode root, int k) {
    if (root == null) return;

    inorder(root.left, k);

    if (count++ == k) {
        res = root.val;
        return;
    }
    
    inorder(root.right, k);
}
```

解法二：

JavaScript Code:

```js

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
function nodeCount(node) {
    if (node === null) return 0;
    
    const l = nodeCount(node.left);
    const r = nodeCount(node.right);
    
    return 1 + l + r;
}
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function(root, k) {
    const c = nodeCount(root.left);
    if (c === k - 1) return root.val;
    else if (c < k - 1) return kthSmallest(root.right, k - c - 1);
    return kthSmallest(root.left, k)
};

```

## 扩展

这道题有一个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?`

大家可以思考一下。