# 【LeetCode】 230. Kth Smallest Element in a BST

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

> 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?

> Constraints:
> * The number of elements of the BST is between `1` to `10^4`.
> * You may assume `k` is always valid, `1 ≤ k ≤ BST's total elements`.

> 給予一個二元搜索樹(BST)，寫一個程式`kthSmallest`可以找出裡面第k小的元素。

> 進階：
> 如果這個BST很常被更改(插入/刪除)，且你也很常需要找到第k小的元素該怎麼辦？你可以優化到常數時間內嗎？

> 限制：
> * BST中的元素介於`1`到`10^4`之間。
> * 你可以假設`k`總是合法，`1 ≤ k ≤ BST's total elements`。

## Example:

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

## Solution

* BST中如果要照大小去遍歷，只要跑中序遍歷即可。
* 因此我們只要跑中序，每次將`k--`，跑到`k == 0`的時候將該節點的值取出即可。
---
* 進階題還沒有研究，先跳過。

### Code

/**
 * 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:
    void InorderTraversal(TreeNode* node, int& k, int& ans)
    {
        if(ans != -1)
            return;
        if(node)
        {
            InorderTraversal(node->left, k, ans);
            k--;
            if(k == 0) ans = node->val;
            InorderTraversal(node->right, k, ans);
        }
    }
    int kthSmallest(TreeNode* root, int k) {
        int ans = -1;
        InorderTraversal(root, k, ans);
        return ans;
    }
};

###### tags: `LeetCode` `C++`