# class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool: # 都是空结点,则当前子树相同,直接返回 true if not...
1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10publicclassSolution {11publicbooleanisSameTree(TreeNode p, TreeNode q) {12if(p ==null&& q ==null)13returntrue;14el...
也可以使用中序遍历的迭代写法,对应之前那道Binary Tree Inorder Traversal,参见代码如下: 解法三: classSolution {public:boolisSameTree(TreeNode* p, TreeNode*q) { stack<TreeNode*>st;while(p || q || !st.empty()) {while(p ||q) {if((p && !q) || (!p && q) || (p->val != q-...
public boolean isSameTree(TreeNode p, TreeNode q) { return inorderTraversal(p,q); } private boolean inorderTraversal(TreeNode p, TreeNode q) { if(p==null&&q==null){ return true; }else if(p==null || q==null){ return false; } //考虑左子树是否符合 if(!inorderTraversal(p.left...
leetcode 100. Same Tree 二叉树DFS深度优先遍历,Giventwobinarytrees,writeafunctiontocheckiftheyareequalornot.Twobinarytreesa
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}". 本题暂用递归解法,具体程序(4ms)如下: 1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {...
100. 相同的树 - 给你两棵二叉树的根节点 p 和 q ,编写一个函数来检验这两棵树是否相同。 如果两个树在结构上相同,并且节点具有相同的值,则认为它们是相同的。 示例 1: [https://assets.leetcode.com/uploads/2020/12/20/ex1.jpg] 输入:p = [1,2,3], q = [1,2,
class Solution { public: bool isSameTree(TreeNode *p, TreeNode *q) { if(p == NULL && q == NULL) return true; else if(p == NULL || q == NULL) return false; bool isleftTreeSame = isSameTree(p->left, q->left); bool isrightTreeSame = isSameTree(p->right,q->right); re...
题目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. 分析 判断两个二叉树是否相等,只要每个都遍历(前,中,后,层次)一下进行对比即可. 笼统的我们也...
Start traversing the tree and each node should return a vector to its parent node. 提示2 The vector should be of length 26 and have the count of all the labels in the sub-tree of this node. 评论(133) 预览评论 💡 讨论区规则