【LeetCode】236. Lowest Common Ancestor of a Binary Tree 解题报告（Python）
标签（空格分隔）： LeetCode
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.me/
题目地址：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
题目描述：
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself
according to the LCA definition.
Note:
- All of the nodes’ values will be unique.
- p and q are different and both values will exist in the binary tree.
题目大意
在一棵普通的二叉树中，找出两个节点的最小公共祖先。
解题方法
这个题竟然没想出来。如果是BST的话，那就很简单了。可以参考一下【LeetCode】235. Lowest Common Ancestor of a Binary Search Tree 解题报告（Java & Python）的做法。
这个题的模式叫做devide and conquer. 如果找到了节点，返回该节点，否则在左右侧分别寻找。然后根据左右侧找到的节点做进一步的判断。如果左右侧分别找到了，那么LCA就是当前节点。否则就在不为空的那边子树里。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if not root or p == root or q == root:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
return left if left else right日期
2018 年 6 月 22 日 ———— 这周的糟心事终于完了