Leetcode之广度优先搜索(BFS)专题-429. N叉树的层序遍历(N-ary Tree Level Order Traversal) 给定一个 N 叉树,返回其节点值的层序遍历。 (即从左到右,逐层遍历)。 例如,给定一个3叉树: 返回其层序遍历: [ [1], [3,2,4], [5,6] ] 说明: 树的深度不会超过1000。 树的节点总数不会超过5000。
【LeetCode】Binary Tree Level Order Traversal 【BFS】 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,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversa...
由于二叉树是从左至右进行输入,故层次遍历通过队列存储每层的结点,它存储的顺序也是前一个结点的左孩子结点、右孩子结点,依次顺序进出队列。 DFS代码参考地址:LeetCode Binary Tree Level Order Traversal 其他题目: Binary Tree Level Order Traversal II 层次遍历从低往root结点输出,如Given binary tree{3,9,20,#...
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...
return its zigzag level order traversal as: [ [3], [20,9], [15,7] ] 1. 2. 3. 4. 5. 题目的意思非常直白。层序遍历整个树,可是第一层正序输出。第二层反序输出,第三层正序输出,以此类推。做法有两种:一、仍然採用level-travel,仅仅是引入一个标记,推断是否反转得到的数列; 二、考虑到stack的特...
LeetCode 102. 二叉树的层序遍历 Binary Tree Level Order Traversal (广度优先搜索(BFS)) 102. 二叉树的层序遍历 给你一个二叉树,请你返回其按 层序遍历 得到的节点值。 (即逐层地,从左到右访问所有节点)。 示例: 二叉树:[3,9,20,null,null,15,7],...
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,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: ...
这个就是二叉树的层序遍历,和leetcode102. Binary Tree Level Order Traversal类似,只是要把最后得到的二维数组逆转一下就成。 C++代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; ...
Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example, given a3-arytree: We should return its level order traversal: [ [1], [3,2,4], [5,6]
这个是在leetcode102. Binary Tree Level Order Traversal的基础上拓展来的,只要加一个判断就行。 C++代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} ...