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）：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def kthSmallest(self, root: TreeNode, k: int) -> int:
        self.inorder_result = []
        
        def inorder(root):
            if root.left:
                inorder(root.left)
                
            self.inorder_result.append(root.val)
            
            if root.right:
                inorder(root.right)
        
        inorder(root)
        
        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。