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# 230.Kth-Smallest-Element-in-a-BST

## 230. Kth Smallest Element in a BST

## 题目地址

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

## 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?
```

## 代码

### Approach #1 Recursion

```java
/**
 * 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) {
        ArrayList<Integer> nums = inorder(root, new ArrayList<Integer>());
    return nums.get(k - 1);
  }

  public ArrayList<Integer> inorder(TreeNode root, ArrayList<Integer> arr) {
    if (root == null)    return arr;
    inorder(root.left, arr);
    arr.add(root.val);
    inorder(root.right, arr);
    return arr;
  }
}
```

### Approach #2 Iteration

```java
class Solution {
  public int kthSmallest(TreeNode root, int k) {
    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();

    while (true) {
      while (root != null) {
        stack.add(root);
        root = root.left;
      }
      root = stack.removeLast();
      if (--k == 0)        return root.val;
      root = root.right;
    }
  }
}
```