题目链接
此题看题解，其实这道题蛮简单的，需要注意二叉搜索树左小右大，为什么还需要看题解，需要反思！
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:self.count=0self.result=Nonedef inorder_traversal(node):if not node:returninorder_traversal(node.left)self.count+=1if self.count==k:self.result=node.valreturninorder_traversal(node.right)inorder_traversal(root)return self.result
)
)
 介绍 】)
