题目链接：
题解
由于是二叉搜索树，两个节点的最近公共祖先就是从 root 找到第一个 val 在两个节点之间的节点。
参考代码
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| #
# @lc app=leetcode.cn id=235 lang=python3
#
# LC235.二叉搜索树的最近公共祖先
#
# @lc code=start
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
pp = min(p.val, q.val)
qq = max(p.val, q.val)
def dfs(rt):
# print(rt.val, pp, qq)
if rt.val >= pp and rt.val <= qq:
return rt
if rt.val > qq:
return dfs(rt.left)
if rt.val < pp:
return dfs(rt.right)
# ans = dfs(root)
return dfs(root)
# @lc code=end
|