LeetCode in Python 236. Lowest Common Ancestor of a Binary Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

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.

 Solution:

# 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
        """
        def isAncestor(m, n):
            if not m:
                return False
            if m == n or m.left == n or m.right == n:
                return True
            return isAncestor(m.left, n) or isAncestor(m.right, n)
        
        if isAncestor(p, q):
            return p
        elif isAncestor(q, p):
            return q
        elif (isAncestor(root.left, p) and isAncestor(root.right, q)) or (isAncestor(root.right, p) and isAncestor(root.left, q)):
            return root
        elif isAncestor(root.left, p) and isAncestor(root.left, q):
            return self.lowestCommonAncestor(root.left, p, q)
        else:
            return self.lowestCommonAncestor(root.right, p, q)        

三种情况：1.p,q有亲缘关系，返回父节点 2.p,q都在同一个子树，递归调用 3.p,q分别在左右子树，LCA是root