Python, LintCode, 88. Lowest Common Ancestor of a Binary Tree

CherryCheekAir

于 2018-05-03 21:10:51 发布

阅读量178

点赞数

CC 4.0 BY-SA版权

分类专栏： LintCode 文章标签： Python LintCode

本文链接：https://blog.youkuaiyun.com/u010342040/article/details/80186701

LintCode 专栏收录该内容

41 篇文章

订阅专栏

本文介绍了一种寻找二叉搜索树中两个节点的最低公共祖先(LCA)的方法。通过递归遍历树结构，当根节点为其中一个目标节点或为空时直接返回该节点或空；若左右子树分别含有目标节点，则当前根节点即为所求。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""


class Solution:
    """
    @param: root: The root of the binary search tree.
    @param: A: A TreeNode in a Binary.
    @param: B: A TreeNode in a Binary.
    @return: Return the least common ancestor(LCA) of the two nodes.
    """
    def lowestCommonAncestor(self, root, A, B):
        # write your code here
        if root == None:
            return None
        if root == A or root == B:
            return root
        leftnode = self.lowestCommonAncestor(root.left, A, B)
        rightnode = self.lowestCommonAncestor(root.right, A, B)
        if leftnode != None and rightnode != None:
            return root
        elif leftnode != None:
            return leftnode
        else:
            return rightnode