leetcode 腾讯精选练习（50 题）236.二叉树的最近公共祖先_the lowest common ancestors of the vertices in a w-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37991005/article/details/94043669

本文探讨了在二叉树中寻找两个节点的最低公共祖先（LCA）的问题，提供了详细的解题思路与代码实现。通过实例说明了如何确定两个节点的最低公共祖先，包括当节点是自身后代的情况。

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原题目

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]

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 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.

思路

第一遍解法

网上好的解法

def lowestCommonAncestor(self, root, p, q):
    if root in (None, p, q): return root
    left, right = (self.lowestCommonAncestor(kid, p, q)
                   for kid in (root.left, root.right))
    return root if left and right else left or right