9-230. 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?

代码实现（他山之石）：

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution:
 9     def kthSmallest(self, root: TreeNode, k: int) -> int:
10         self.inorder_result = []
11         
12         def inorder(root):
13             if root.left:
14                 inorder(root.left)
15                 
16             self.inorder_result.append(root.val)
17             
18             if root.right:
19                 inorder(root.right)
20         
21         inorder(root)
22         
23         return self.inorder_result[k-1]

分析：

通过inorder函数，从根节点开始，找到最左下角的元素，然后从小到大访问树的每一个节点（并记录每个当前节点的值，存储在inorder_result中的值就是从小到大排列的），直至遍历所有节点。

最终inorder_result中第k-1个位置上的值就是第k小的数。

9-230. Kth Smallest Element in a BST

原文：https://www.cnblogs.com/tbgatgb/p/10941272.html