# 235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先

@TOC

[LeetCode]

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/open in new window

Total Accepted: 67533 Total Submissions: 178900 Difficulty: Easy

## # 题目描述

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]

``````        _______6______
/              \
___2__          ___8__
/      \        /      \
0      _4       7       9
/  \
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.
``````

Note:

1. All of the nodes' values will be unique.
2. p and q are different and both values will exist in the BST.

## # 解题方法

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||p==root||q==root){
return root;
}
//devide
TreeNode left=lowestCommonAncestor(root.left,p,q);
TreeNode right=lowestCommonAncestor(root.right,p,q);

//conquer
if(left!=null&&right!=null){
return root;
}else if(left!=null){
return left;
}else{
return right;
}
}
}
``````

AC:11ms

``````# 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 min(p.val, q.val) <= root.val and max(p.val, q.val) >= root.val:
return root
elif p.val < root.val and q.val < root.val:
return self.lowestCommonAncestor(root.left, p, q)
elif p.val > root.val and q.val > root.val:
return self.lowestCommonAncestor(root.right, p, q)
``````

``````# 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
"""
while root:
if p.val < root.val and q.val < root.val:
root = root.left
elif p.val > root.val and q.val > root.val:
root = root.right
else:
break
return root
``````

``````# 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 root == p or root == q:
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

2016/5/1 14:05:25 2018 年 6 月 22 日 ———— 这周的糟心事终于完了 2018 年 11 月 19 日 —— 周一又开始了