其实题目的意思就是相当于二叉树每一行的最右边的一个元素,那么简单,先bfs后取每一个数组的最后一位数就可以了,代码如下: 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(NU...
TreeNode*node =q.front(); q.pop();if(node->left) q.push(node->left);if(node->right) q.push(node->right); } }returnres; } }; LeetCode All in One 题目讲解汇总(持续更新中...)
问题链接 英文网站:199. Binary Tree Right Side View中文网站:199. 二叉树的右视图问题描述Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the n…
题目地址:https://leetcode.com/problems/binary-tree-right-side-view/description/ 题目描述: Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example: Given the following binary tree, 1 <---...
LeetCode199题 Binary Tree Right Side View, 解题思路。 1、先读题,凡是同一层的,有右边的节点,只添加右边的。 2、用bfs去解决,队列先添加右子树,再添加左子树,队列除了带node 信息,还得有当前层数的信息。 3、循环处理,当前层数没有被使用,就添加node的val。
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example: Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- ...
Blogger:https://blog.baozitraining.org/2019/10/leetcode-solution-199-binary-tree-right.html Youtube:https://youtu.be/_g6pN64bF-o 博客园: https://www.cnblogs.com/baozitraining/p/11595617.html B站: https://www.bilibili.com/video/av69112013/ ...
01.Binary Tree Inorder Traversal.wmv 02.Path Sum.wmv 03.Maximum Depth of Binary Tree.wmv 55.LeetCode 刷题 – 常用算法思想 笔记和源码 55. LeetCode 刷题 – 常用算法思想.mht 01.贪心算法思想 .wmv 02.回溯算法思想.wmv 03.动态规划算法思想.wmv 56.OpenCV – 安装和简介 笔记和源码 opencv需要的...
http://www.programcreek.com/2014/04/leetcode-binary-tree-right-side-view-java/ 看了这篇博客就懂了。 ** 总结: queue来遍历tree ** Anyway, Good luck, Richardo! My code: /** * Definition for a binary tree node. * public class TreeNode { ...
Given a binary tree, imagine yourself standing on therightside of it, return the values of the nodes you can see ordered from top to bottom. 一颗二叉树,获取从右边看去能看到的每层的第一个节点,就是每层最右侧的节点。 广度优先遍历,做一点小修改就行: ...