# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def rightSideView(self, root): """ :type root: TreeNode :rtype: List[int]…

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 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

原题地址:https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ 题意: Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6. 解题思路:这道题是在树中寻找一条路…

好久不写了,最近忙毕业论文呢. 这个题,就是说一个二叉树,你从右边看,你能看到的数有哪些(会被遮挡) 其实抽象出来就是说...二叉树每层最右边的数有哪些.. 那我们按层遍历一次就好了. /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} *…

题意: 给一棵二叉树,要求收集每层的最后一个节点的值.按从顶到底装进vector返回. 思路: BFS比较简单,先遍历右孩子就行了. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solut…

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def levelOrder(self, root): """ :type root: TreeNode :rtype: List[List[int…

[LeetCode]199. Binary Tree Right Side View 解题报告(Python) 标签: LeetCode 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 no…

// 我的代码 package Leetcode; /** * 199. Binary Tree Right Side View * address: https://leetcode.com/problems/binary-tree-right-side-view/ * Given a binary tree, imagine yourself standing on the right side of it, * return the values of the nodes you can…

leetcode 199. Binary Tree Right Side View 这个题实际上就是把每一行最右侧的树打印出来,所以实际上还是一个层次遍历. 依旧利用之前层次遍历的代码,每次大的循环存储的是一行的节点,最后一个节点就是想要的那个节点 class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> result; if(root == NULL) return resul…

Leetcode之深度优先搜索(DFS)专题-199. 二叉树的右视图(Binary Tree Right Side View) 深度优先搜索的解题详细介绍,点击 给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值. 示例: 输入: [1,2,3,null,5,null,4] 输出: [1, 3, 4] 解释: 1 <--- / \ 2 3 <--- \ \ 5 4 <--- 分析:把每一行的值保存起来,最后再把每一行最后一个放进ans里. AC代码…