NewsType结构: Id ParentId Name children(List<NewsType>) public void LoopToAppendChildren(List<NewsType> all, NewsType curItem) { var subItems = all.Where(ee => ee.ParentId==curItem.Id).ToList(); curItem.children = new List<NewsType>();

子组件  given-person.html <!--权限设置-选择员工--> <li [class.noborder]="!dir.shierarchy" *ngFor="let dir of directories" [ngStyle]="{'margin-left': 5+(dir.shierarchy*.5)+'px'}"> <span *ngIf="dir.employee.length &

function loop(data) { let office = data.map(item => { if(item.type == '1' ||item.type == '2') { item = {...item,disabled:true,children:loop(item.children)} return item }else { return item } }) return office } let officeDataTree = loop(officeData)

第1章 零基础转CS,如何准备? · 转专业找CS工作怎么办? · 零基础如何在最短时间内拿到offer? · 如何写好简历? · IT技术面试内容有哪些? · JAVA语言怎么入门? 第2章 数组与循环 Array & Loops · Java语言基础 I 1) IDE介绍:IntelliJ 2) int变量及其范围 3) char变量,什么是Unicode 4) boolean变量的与或非运算 5) 如何写好if语句 6) for循环和while循环 · 实战面试真题 1) characte

http://acm.uestc.edu.cn/#/problem/show/1324 卿学姐与公主 Time Limit: 2000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)     某日,百无聊赖的卿学姐打开了某11区的某魔幻游戏 在这个魔幻的游戏里,生活着一个美丽的公主,但现在公主被关押在了魔王的城堡中. 英勇的卿学姐拔出利刃冲向了拯救公主的道路. 走过了荒野,翻越了高山,跨过了大洋,卿学姐来到了魔

树的结构 树(tree)是一种抽象数据类型或是实现这种抽象数据类型的数据结构,用来模拟具有树状结构性质的数据集合 它具有以下的特点: ①每个节点有零个或多个子节点: ②没有父节点的节点称为根节点: ③每一个非根节点有且只有一个父节点: ④除了根节点外,每个子节点可以分为多个不相交的子树: 树的分类 二叉树 二叉树:每个节点最多含有两个子树的树称为二叉树. 二叉树中一些专业术语: 父节点:A节点就是B节点的父节点,B节点是A节点的子节点 兄弟节点:B.C这两个节点的父节点是同一个节点,所以他们互称

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=48 Tree Summing  Background LISP was one of the earliest high-level programming languages and, with FORTRAN, is one of the olde

; i < dr.Rows.Count; i++)             {                 )                 {                     n.id = Convert.ToInt32(dr.Rows[i][; i < dr.Rows.Count; i++)             {                 tree n = ;             BindNew(model);];                 return

hdu5044 Tree 树链拆分.点细分.刚,非递归版本 //#pragma warning (disable: 4786) //#pragma comment (linker, "/STACK:16777216") //#pragma comment(linker, "/STACK:60400000,60400000") //HEAD #include <cstdio> #include <ctime> #include <cstd

Given an n-ary tree, return the postorder traversal of its nodes' values. For example, given a 3-ary tree: Return its postorder traversal as: [5,6,3,2,4,1]. Note: Recursive solution is trivial, could you do it iteratively? ---------------------------

Given an n-ary tree, return the preorder traversal of its nodes' values. For example, given a 3-ary tree: Return its preorder traversal as: [1,3,5,6,2,4]. Note: Recursive solution is trivial, could you do it iteratively? -----------------------------

Given a n-ary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. For example, given a 3-ary tree: We should return its max depth, which is 3. Note: The dept

Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less thanthe node's key. The right subtree of a node contains only nodes with keys 

Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? --------------------------------------------------

Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? ----------------------------------------------------

Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. Example 1: Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1] Output: [1,2,3,4,5,6,7] Note: 1 <=

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2

Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? ---------------------------------------------------

144. Binary Tree Preorder Traversal Difficulty: Medium Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it it

首先我有必要记录下来这段代码,因为我遇到了这个问题, 然后没有解决 后来, 前段说我找到一段代码给我看看, 我并没有在意, 然后她实现了, 她实现了,她真的实现了, 我... 为了感谢她,我陪她玩了一中午的五子棋, 并假装输了几把 先说目的: 递归迭代 再看数据: 其实有个前提, 就是aid不能重复 fid是父级id, 指向的aid 如果 aid == fid , 证明fid的那个aid是子集 如果 fid == 0 就证明是顶级 代码: def aaa(oldArr, fid): newArr