[LeetCode] Level Order Traversal 题目说明 Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3]...
LeetCode | Binary Tree Level Order Traversal Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree{3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3...
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 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...
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...
15 7 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】
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]] ...
103. Binary Tree Zigzag Level Order Traversal.md node = nodes[i] values.append(node.val) first, second = access_orders[level % 2] first_node = getattr(node, first) nodchip Feb 27, 2024 リフレクション的な機能は極力使用しないほうが良いと思います。理由は、コンパイル時の静...