ICode9

精准搜索请尝试: 精确搜索
  • 5084. Insufficient Nodes in Root to Leaf Paths2019-06-09 14:53:39

    Given the root of a binary tree, consider all root to leaf paths: paths from the root to any leaf.  (A leaf is a node with no children.) A node is insufficient if every such root to leaf path intersecting this node has sum strictly less than

  • Leetcode-5084 Insufficient Nodes in Root to Leaf Paths(根到叶路径上的不足节点)2019-06-09 12:53:32

    这题出的不好 1 class Solution 2 { 3 public: 4 int go(TreeNode* root, int limit,int k) 5 { 6 if(!(root->left) && !(root->right)) 7 { 8 int t = k+root->val; 9 if(t <

  • [Swift]LeetCode1080. 根到叶路径上的不足节点 | Insufficient Nodes in Root to Leaf Paths2019-06-09 12:40:05

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)➤GitHub地址:https://www.cnblogs.com/strengthen/p/10993155.html ➤原文地址:https://www.cnbl

  • 有根树的表达 Aizu - ALDS1_7_A: Rooted Trees2019-06-07 19:49:11

    有根树的表达 题目:Rooted Trees Aizu - ALDS1_7_A  A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).  Fig. 1

  • leetcode 129. Sum Root to Leaf Numbers2019-05-31 13:39:16

    https://www.cnblogs.com/grandyang/p/4273700.html sum代表当前节点的和。 这个题要从上往下的思路去做。 class Solution {public: int sumNumbers(TreeNode* root) { return sumNumbers(root,0); } int sumNumbers(TreeNode* root,int sum){ if(!root)

  • [LeetCode] Leaf-Similar Trees 叶结点相似的树2019-04-25 23:41:05

    Consider all the leaves of a binary tree.  From left to right order, the values of those leaves form a leaf value sequence. For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8). Two binary trees are considered leaf-similar

  • Leetcode-5017 Sum of Root To Leaf Binary Numbers(从根到叶的二进制数之和)2019-04-07 13:38:00

    1 typedef long long ll; 2 class Solution 3 { 4 public: 5 ll rnt = 0; 6 void go(TreeNode* root,int val) 7 { 8 val = (val*2+root->val)%1000000007; 9 if(root->left==NULL && root->right==NU

  • LeetCode-Python-872. 叶子相似的树2019-04-05 19:53:51

    请考虑一颗二叉树上所有的叶子,这些叶子的值按从左到右的顺序排列形成一个 叶值序列。 举个例子,如上图所示,给定一颗叶值序列为 (6, 7, 4, 9, 8) 的树。 如果有两颗二叉树的叶值序列是相同,那么我们就认为它们是 叶相似 的。 如果给定的两个头结点分别为 root1 和 root2

  • vxlan二层互通-有隧道方式2019-03-18 14:56:10

    一、VXLAN基本二、实验拓扑leaf-1A和Leaf-1B部署VXLAN,Leaf-1a和leaf-1b两台交换机静态创建VXLAN隧道,实现同网段PC1和PC2相互通信配置思路:1、spine1 leaf-a leaf-b配置loopback口,作为vtep ip2、spine1 leaf-1 leaf-b配置ospf协议3、spine1作为transit,无需配置vxlan,按照正常ip转发v

  • LintCode 481 Binary Tree Leaf Sum2019-03-17 13:53:05

    思路 一道巨简单的题目,单纯想练习一下递归。 复杂度 时间复杂度O(n) 空间复杂度最坏情况O(n) 代码 /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * th

  • PAT (Advanced Level) Practice 1119 Pre- and Post-order Traversals 插入法构建二叉树2019-03-05 16:52:59

    一、概述 根据前序和后序遍历,判断二叉树是否唯一。 乍一看有点懵,怎么判断呢?还是要上手模拟一下。 前序遍历结构:根节点、左子树、右子树 后序遍历结构:左子树、右子树、根节点 假设两个树都不空,那么是可以唯一确定一棵树的:通过前序确定左子树的根,后序确定右子树的根,然后分别遍历,

  • LeetCode题解之Sum Root to Leaf Numbers2019-03-02 19:42:46

    1、题目描述 2、问题分析 记录所有路径上的值,然后转换为int求和。   3、代码 1 vector<string> s; 2 int sumNumbers(TreeNode* root) { 3 traverse(root, ""); 4 int sum = 0; 5 for(auto it = s.begin(); it != s.end(); it++) { 6

  • LeetCode题解之Leaf-Similar Trees2019-02-24 14:50:17

    1、题目描述 2、问题分析 将叶子节点的值放入vector,然后比较。   3、代码 1 bool leafSimilar(TreeNode* root1, TreeNode* root2) { 2 vector<int> v1; 3 vector<int> v2; 4 5 findLeaf(root1, v1); 6 findLeaf(root2, v2); 7

  • 机器学习笔记(五):决策树理论与简单使用2019-02-24 10:55:16

    1.决策树构造 难点:如何构造树 选择什么样的节点做根节点 2.熵 :衡量标准,用于选择哪个节点做当前的节点 表示随机变量的不确定性(物体内部的混乱程度【所以熵越小越好】) ID3算法: 就是基与信息增益的算法 问题: 假设以ID(每条数据都是不同的 或者 那些分的很细但是没有意义的

  • LeetCode 872 Leaf-Similar Trees 解题报告2019-02-17 09:40:48

    题目要求 Consider all the leaves of a binary tree.  From left to right order, the values of those leaves form a leaf value sequence. Two binary trees are considered leaf-similar if their leaf value sequence is the same. Return true if and only if the t

  • Christmas Spruce2019-02-13 09:45:19

      Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to 

  • CSE 491 Introduction to Bioinformatics2019-02-12 19:47:49

    代写CSE 491留学生作业、Java,c++程序作业代写、代做Python,Java实验作业、代做Bioinformatics作业CSE 491 Introduction to BioinformaticsHomework 2Assigned on February 5, 2019. Written solution due at start of class on February 14, 2019.Type-set solutions are preferred

  • [codeforces]Codeforces Global Round 1 F. Nearest Leaf2019-02-08 08:54:04

    题解:  语文题????  上面说的一段代码 告诉你的是  节点编号顺序与dfs序顺序一致  也就是你  dfs序以后编号就是[1,n]  根据这个特性  那么我们只需要维护每个叶子节点到查询v的距离即可  那么我们只需要离线所有查询 然后对子树修改即可   用线段树来维护区间加和区间最

  • LeetCode-129-Sum Root to Leaf Numbers2019-02-05 19:47:42

    算法描述: Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum of all root-to-leaf numbers. Note: A leaf is

  • [Swift Weekly Contest 122]LeetCode988. 从叶结点开始的最小字符串 | Smallest String Starting From Leaf2019-02-04 13:47:52

    Given the root of a binary tree, each node has a value from 0 to 25 representing the letters 'a' to 'z': a value of 0 represents 'a', a value of 1 represents 'b', and so on. Find the lexicographically smallest strin

专注分享技术,共同学习,共同进步。侵权联系[81616952@qq.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有