LeetCode 235. Lowest Common Ancestor of a Binary Search Tree
Description
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
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 binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
描述
给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]
示例 1:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6
解释: 节点 2 和节点 8 的最近公共祖先是 6。
示例 2:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
思路
- 根据二叉搜索树的性质， lowest common ancestor 节点的值一定在p和q之间.
- 如果当前节点的值大于p和q节点的值，说明LCA在当前节点左边.
- 如果当前节点的值小于p和q节点的值，说明LCA在当前节点右边.
# -*- coding: utf-8 -*-
# @Author: 何睿
# @Create Date: 2019-02-01 18:26:43
# @Last Modified by: 何睿
# @Last Modified time: 2019-02-01 18:44:19
# 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, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
# 根据二叉搜索树的性质， lowest common ancestor 节点的值一定在p和q之间
# 如果当前节点的值大于p和q节点的值，说明LCA在当前节点左边
# 如果当前节点的值小于p和q节点的值，说明LCA在当前节点右边
if p.val < q.val:
return self._recursion(root, p, q)
else:
return self._recursion(root, q, p)
def _recursion(self, root, p, q):
# p节点的值一定小于q节点的值
if p.val <= root.val <= q.val: return root
if root.val < p.val and root.val < q.val:
return self._recursion(root.right, p, q)
if root.val > p.val and root.val > q.val:
return self._recursion(root.left, p, q)
