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...
/** * 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: bool isSameTree(TreeNode* p, TreeNode* q) { if(!p && !q) { ...
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
voidserialize(TreeNode* root,string& str){if(root ==nullptr) {str = str +"#";}else{str = str +","+ to_string(root->val); serialize(root->left, str);serialize(root->right, str);}} 可以看到,该代码实际上还是二叉树的遍历。
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{...