码迷,mamicode.com
首页 > 其他好文 > 详细

#Leetcode# 236. Lowest Common Ancestor of a Binary Tree

时间:2018-12-24 00:02:14      阅读:131      评论:0      收藏:0      [点我收藏+]

标签:tor   div   代码   its   style   ref   upload   different   amp   

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]

技术分享图片

 

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.

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root || p == root || q == root) return root;
        TreeNode *left = lowestCommonAncestor(root -> left, p, q);
        TreeNode *right = lowestCommonAncestor(root -> right, p, q);
        if(left && right) return root;
        return left ? left : right;
    }
};

  呜呜呜 今天在 PAT 甲级接连碰壁 总是差一点 AC 啊  刚开始看这个题似曾相识 仔细翻一下才发现上一个是二叉搜索树这个只是普通二叉树

#Leetcode# 236. Lowest Common Ancestor of a Binary Tree

标签:tor   div   代码   its   style   ref   upload   different   amp   

原文地址:https://www.cnblogs.com/zlrrrr/p/10165813.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!