2019-03-28 15:25:43 问题描述: 问题求解: 写过的最好的Hard题之一. 初看本题,很经典的路径和嘛,dfs一遍肯定可以得到某个节点到其他所有节点的距离和.这种算法的时间复杂度是O(n ^ 2).看一下数据量,emmm,果然不行.这个数据量一看就知道只能是O(n)的算法了. 只遍历一遍最多只能得到一个解,因此本题肯定是需要遍历至少两遍的. 在第一遍遍历的时候我们需要保存下两个值,一个是当前节点的subtree的路径总和,一个是当前节点的subtree的总的节点数. 在第二遍遍…

Sum of Distances in Tree An undirected, connected tree with N nodes labelled 0...N-1 and N-1 edges are given. The ith edge connects nodes edges[i][0] and edges[i][1] together. Return a list ans, where ans[i] is the sum of the distances between node i…

An undirected, connected tree with N nodes labelled 0...N-1 and N-1 edges are given. The ith edge connects nodes edges[i][0]and edges[i][1] together. Return a list ans, where ans[i] is the sum of the distances between node i and all other nodes. Exam…

An undirected, connected tree with N nodes labelled 0...N-1 and N-1 edges are given. The ith edge connects nodes edges[i][0] and edges[i][1] together. Return a list ans, where ans[i] is the sum of the distances between node i and all other nodes. Exa…

LeetCode刷题记录 传送门 Description An undirected, connected treewith N nodes labelled 0...N-1 and N-1 edges are given. The ith edge connects nodes edges[i][0] and edges[i][1] together. Return a list ans, where ans[i] is the sum of the distances between nod…

There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given the integer n and the array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. Return an array…

思路: 树形dp. 实现: class Solution { public: void dfs(int root, int p, vector<vector<int>>& G, vector<int>& cnt, vector<int>& res) { for (auto it: G[root]) { if (it == p) continue; dfs(it, root, G, cnt, res); cnt[root] += cnt…

Path Sum leetcode java 描述 Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13…

CodeChef Sum of distances(分治) 题目大意 有一排点,每个点 i 向 \(i + 1, i + 2, i + 3\) 分别连价值为 \(a_i,b_i,c_i\) 的有向边,问两两间最短路之和 数据范围 \(1 \le n \le 10^5\) 解题思路 这种题已经从新颖变成套路了(唉) 考虑分治,很容易发现如果断掉连续的三个点那么图就不再联通,我们从中间找三个点,然后分别向两边跑最短路,设点 i 到三点最短距离为 \(x_1,x_2,x_3\),三点到 j 最短距离为…

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree andsum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 return true…

http://www.itint5.com/oj/#13 要注意,一是空路径也可以,所以最小是0.然后要时刻注意路径顶多有两条子路径+根节点组成,所以更新全局最值时和返回上一级的值要注意分清. #include <climits> using namespace std; int maxPathHelper(TreeNode *root, int &max) { if (root == NULL) { return 0; } int root_val = root->val; /…

题目如下: Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on. Return the smallest level X such that the sum of all the values of nodes at level X is maximal. Example 1: Input: [1,7,0,7,-8,null,null] Out…

""" BFS遍历题,一遍AC Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on. Return the smallest level X such that the sum of all the values of nodes at level X is maximal. Example 1: Input: [1,7,0…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 BFS 日期 题目地址:https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/ 题目描述 Given the root of a binary tree, the level of its root is 1, the level of its children is 2, a…

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the nod…

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. For exampl…

［抄题］: 给一个二叉查找树以及一个节点,求该节点的中序遍历后继,如果没有返回null ［思维问题］: 不知道分合算法和后序节点有什么关系:直接return表达式就行了,它自己会终止的. ［一句话思路］: 比root大时直接扔右边递归,比root小时 考虑是左边递归还是就是root ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 要定义left节点,留着做后续的比较 ［二刷］: ［三刷］: ［四刷］: ［五刷］: ［总结］…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

DFS基础 深度优先搜索(Depth First Search)是一种搜索思路,相比广度优先搜索(BFS),DFS对每一个分枝路径深入到不能再深入为止,其应用于树/图的遍历.嵌套关系处理.回溯等,可以用递归.堆栈(stack)实现DFS过程. 关于广度优先搜索(BFS)详见:算法与数据结构基础 - 广度优先搜索(BFS) 关于递归(Recursion)详见:算法与数据结构基础 - 递归(Recursion) 树的遍历 DFS常用于二叉树的遍历,关于二叉树详见: 算法与数据结构基础 - 二叉查找树…

[94]Binary Tree Inorder Traversal [95]Unique Binary Search Trees II (2018年11月14日,算法群) 给了一个 n,返回结点是 1 - n 的所有形态的BST. 题解:枚举每个根节点 r, 然后递归的生成左右子树的所有集合,然后做笛卡尔积. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * Tr…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

# Title Solution Acceptance Difficulty Frequency     4 Median of Two Sorted Arrays       27.2% Hard     10 Regular Expression Matching       25.6% Hard     23 Merge k Sorted Lists       35.8% Hard     25 Reverse Nodes in k-Group       37.7% Hard    …

Contest 81 (2018年11月8日,周四,凌晨) 链接:https://leetcode.com/contest/weekly-contest-81 比赛情况记录:结果:3/4, ranking: 440/2797.这次题目似乎比较简单,因为我比赛的时候前三题全做出来了(1:12:39),然后第四题有思路,正在写,没写完,比赛完了写完提交也对了. [821]Shortest Distance to a Character(第一题 4分) 给了一个单词(字符串)s,和单词中的任意一个字母…

[98]Validate Binary Search Tree [99]Recover Binary Search Tree [100]Same Tree [101]Symmetric Tree [104]Maximum Depth of Binary Tree [105]Construct Binary Tree from Preorder and Inorder Traversal [106]Construct Binary Tree from Inorder and Postorder T…

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22, 5 / \ 4 8…

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 给定一个二叉树和一个和,判断这个树中是否有一个从根到叶子的路径,使其这个路径上面的所有节点值的和为这个给定的值. 例如: 给定下面的二叉树,并且和为22. 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 返回true,因为这里面存在一个根到叶子的路径 5->4->11->2,使其他们的和为22. ++…

题目: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22, 5 / \…

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 return true…

题目: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree and  sum = 22 , 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 retu…

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children. Example: Given the below binary tree and sum = 22, / \ / / \…