leetcode---Same Tree Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 初步构想:二叉树判断相等问题,肯定是先判断左子树再判断右子树哇,但是,树空!!! 注意事项...
classSolution {public:boolisSameTree(TreeNode* p, TreeNode*q) { stack<TreeNode*>st1, st2; TreeNode*head1, *head2;while(p || q || !st1.empty() || !st2.empty()) {while(p ||q) {if((p && !q) || (!p && q) || (p->val != q->val))returnfalse; st1.push(p); s...
* struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if(!p && !q) { return true; } else if(p && q) { return ...
通过的Java代码如下: /*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/classSolution{publicbooleanisSameTree(TreeNodep,TreeNodeq){if(p==null&&q==null){returntrue;}elseif(p==null&&q!
leetcode笔记:Same Tree 一. 题目描写叙述 Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 二. 题目分析...
Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 思路:太简单! AI检测代码解析 boolisSameTree(TreeNode *p, TreeNode *q) {if(p == NULL && q == NULL...
1519. 子树中标签相同的节点数 - 给你一棵树(即,一个连通的无环无向图),这棵树由编号从 0 到 n - 1 的 n 个节点组成,且恰好有 n - 1 条 edges 。树的根节点为节点 0 ,树上的每一个节点都有一个标签,也就是字符串 labels 中的一个小写字符(编号为 i 的 节点的标签就
这个题目是leetcode的第572题,要求是这样的:给定两颗二叉树A和B,判断B是否是A的子树。 在下面这个例子中可以看到B是A的子树。 想一想该怎样解决这个问题呢? 如果B是A的一颗子树,那么B一定和A的一个颗子树完全一样,因此我们可以实现一个函数isSame来判断两颗二叉树是否完全相同,这个函数非常容易实现: ...
题目链接[https://leetcode-cn.com/problems/same-tree/]tag: Easy; DFS; BFS; question: Given ...
same-tree //先序遍历递归classSolution{public:boolisSameTree(TreeNode*p,TreeNode*q){if(!p&&!q)returntrue;if(!p||!q)returnfalse;returnp->val==q->val&&isSameTree(p->left,q->left)&&isSameTree(p->right,q->right);}};//后序遍历递归,会比先序遍历的快,因为从下到上classSolution{...