
# 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:
#当前节点为空了 直接返回None;当前节点为p 或 q 则找到了它们自身 返回给上一层做判断
if not root or root.val == p.val or root.val == q.val: return root
#递归左子树和右子树
l = self.lowestCommonAncestor(root.left, p , q)
r = self.lowestCommonAncestor(root.right, p , q)
##两边都有 证明p q一定是分别在左右节点上 root一定是公共祖先;只有一边有 root一定不是最近公共祖先 直接把子节点传递回来
return root if l and r else l or r
本文来自博客园,作者:topass123,转载请注明原文链接:https://www.cnblogs.com/topass123/p/12764706.html
浙公网安备 33010602011771号