problem 965. Univalued Binary Tree 参考 1. Leetcode_easy_965. Univalued Binary Tree; 完…

题目来源: https://leetcode.com/problems/univalued-binary-tree/submissions/ 自我感觉难度/真实难度: 题意: 分析: 自己的代码: class Solution(object): def isUnivalTree(self, root): """ :type root: TreeNode :rtype: bool """ return self.dsf(root.val,root)…

A binary tree is univalued if every node in the tree has the same value. Return true if and only if the given tree is univalued. Example 1: Input: [1,1,1,1,1,null,1] Output: true Example 2: Input: [2,2,2,5,2] Output: false Note: The number of nodes i…

题目要求 A binary tree is univalued if every node in the tree has the same value. Return true if and only if the given tree is univalued. 题目分析及思路 题目要求当输入的二叉树的每个结点都有相同的值则返回true,否则返回false.可以使用队列保存每个节点,用val保存root节点的值.如果弹出的数字不等于val不等于root节点就立刻返回false.如果全部判断完…

A binary tree is univalued if every node in the tree has the same value. Return true if and only if the given tree is univalued. 太简单了,签到题.…

题目如下: A binary tree is univalued if every node in the tree has the same value. Return true if and only if the given tree is univalued. Example 1: Input: [1,1,1,1,1,null,1] Output: true Example 2: Input: [2,2,2,5,2] Output: false Note: The number of n…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 BFS DFS 日期 题目地址:https://leetcode.com/problems/univalued-binary-tree/ 题目描述 A binary tree is univalued if every node in the tree has the same value. Return true if and only if th…

A binary tree is univalued if every node in the tree has the same value. Return true if and only if the given tree is univalued. Example 1: Input: [1,1,1,1,1,null,1] Output: true Example 2: Input: [2,2,2,5,2] Output: false Note: The number of nodes i…

这是悦乐书的第366次更新,第394篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第228题(顺位题号是965).如果树中的每个节点具有相同的值,则二叉树是单一的.当且仅当给定树是单一时才返回true. 1 / \ 1 1 / \ \ 1 1 1 输入: [1,1,1,1,1,null,1] 输出: true 2 / \ 2 2 / \ 5 2 输入: [2,2,2,5,2] 输出: false 注意: 给定树中的节点数量将在[1,100]范围内. 每个节点的值将是…

1.题目描述 2.问题分析 遍历一遍树,然后将所有节点的数值放入到一个set中,最后检查set中元素的个数是否为1. 3.代码 bool isUnivalTree(TreeNode* root) { set<int> s; preOrder(root, s); ; } void preOrder(TreeNode* root, set<int> &s) { if (root == NULL) return; s.insert(root->val); preOrder(…