代码 publicArrayList<ArrayList<Integer>>levelOrder(TreeNode root) {//Note: The Solution object is instantiated only once and is reused by each test case.ArrayList<ArrayList<Integer>> ans=newArrayList<ArrayList<Integer>>();if(root==null)returnans; Queue<TreeNode> list=newLinkedList<TreeNode>();...
return its level order traversal as: [ [3], [9,20], [15,7] ] 这道题利用宽度优先搜索就可以了,具体程序(8ms)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), ri...
leetcode 107. Binary Tree Level Order Traversal II Given a binary tree, return thebottom-up level ordertraversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree[3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 1...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript 代码运行次数:0 运行 AI代码解释 3/\920/\157 return its level order traversal as: 代码...
Leetcode: Binary Tree Level Order Traversal II 题目: Given a binary tree, return thebottom-up level ordertraversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7},...
Can you solve this real interview question? Binary Tree Level Order Traversal - Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: [https://assets.leetcode.c
这道题和LeetCode笔记:107. Binary Tree Level Order Traversal II是姊妹题,解题思路都是一样的,只是结果要求的顺序是反的,同样有两种方法,也就是经常说到的DFS深度优先遍历和BFS广度优先遍历。 BFS: 广度优先遍历就是一层层地攻略过去,把每一层的所有节点都记录下来再走向下一层。因为每层会有多个节点,不是简...
return its level order traversal as: [ [3], [9,20], [15,7] ] 【Idea】 这个题也不难, 因为O(n)还蛮能打的, 所以po一下 BFS要更好写一些。 这里每层遍历node时直接用了for,没有用queue,不然还要另起个变量标记 【Solution】 # Definition for a binary tree node.# class TreeNode:# def ...
hroc135/leetcode#26 olsen-blue/Arai60#27 103. Binary Tree Zigzag Level Order Traversal 58b6ace olsen-blue reviewed Mar 2, 2025 View reviewed changes 103/step3.cpp vector<TreeNode*> current_level_nodes = {root}; while (!current_level_nodes.empty()) { vector<TreeNode*> next_lev...
给你一个二叉树,请你返回其按层序遍历(即逐层地,从左到右访问所有节点)的结果。 具体说明如下: 给定一个二叉树的根节点root,请你返回其节点值的层序遍历(逐层遍历)。 每层的节点值应作为一个列表放在最终的结果列表中。 3/\920/\157## 遍历结果[[3],[9,20],[15,7]] ...