An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

PAT甲级1123. Is It a Complete AVL Tree 题意: 在AVL树中,任何节点的两个子树的高度最多有一个;如果在任何时候它们不同于一个,则重新平衡来恢复此属性.图1-4说明了旋转规则. 现在给出一系列插入, 您应该输出生成的AVL树的级别遍历序列,并告知它是否是完整的二叉树. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行包含正整数N(<= 20). 一行中的所有数字都以空格分隔. 输出规格: 对于每个测试用例,将键逐个插入到初始空的AVL树中.然后首先…

嫌排版乱的话可以移步我的CSDN:https://blog.csdn.net/weixin_44385565/article/details/89390802 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by…

题目 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node difer by at most one; if at any time they difer by more than one, rebalancing is done to restore this property. Figures 1-4 illus…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805351302414336 题意: 给定n个树,依次插入一棵AVL树,按照层序遍历输出,最后判断这棵AVL树是不是完全二叉树. 思路: 这道题过段时间还要再来手搓一发.AVL模板要记住. 判断是不是完全二叉树的话只用看,如果有一个节点儿子是空,而他之后又出现了至少有一个儿子的节点的话,就不是完全二叉树.[蛮巧妙的] #include<cstdio> #inclu…

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6806292.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 给一个序列,对其进行AVL树的插入操作然后输出该二叉树的层次遍历序列若该二叉树为满二叉树,则输出YES,否则输出NO. AVL树的插入操作,模板题,不多说了.可以在BFS的同时,判断其是否为满二叉树.一层层遍历,每遍历一个节点,cnt++,统计节点个数.当第…

AVL树的插入,旋转. #include<map> #include<set> #include<ctime> #include<cmath> #include<queue> #include<string> #include<stack> #include<vector> #include<cstdio> #include<cstring> #include<iostream&g…

题意: 输入一个正整数N(<=100),接着输入N行每行包括0~N-1结点的左右子结点,接着输入一行N个数表示数的结点值.输出这颗二叉排序树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; pair<]; ]; ; ]; void dfs(int x){ ) return ; dfs(a[x].first); ans[x]=b[++c…

Source: PAT A1123 Is It a Complete AVL Tree (30 分) Description: An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one,…

1123 Is It a Complete AVL Tree(30 分) An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to rest…

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illust…

题意:给定结点个数n和插入序列,判断构造的AVL树是否是完全二叉树? 思路:AVL树的建立很简单.而如何判断是不是完全二叉树呢?通过层序遍历进行判断:当一个结点的孩子结点为空时,则此后就不能有新的结点入队.若没有,则是完全二叉树,否则不是. 代码: #include <cstdio> #include <algorithm> #include <iostream> #include <vector> #include <queue> using…

题目描述 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.Given any two nodes in a binary tree, you are supposed to find their LCA. 最小共同祖先(LCA)是一棵树中两个节点U和V最深的那个公共父节点.要求给一棵树,以及两个节点,请你…

https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than th…

1043 Is It a Binary Search Tree (25 分)   A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a…

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with…

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contains one test case. For each case, th…

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with keys greate…

题意: 输入一个正整数N(<=1000),接着输入N个整数([-1000,1000]),依次插入一棵初始为空的二叉排序树.输出最底层和最底层上一层的结点个数之和,例如x+y=x+y. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct Node{ int value; int vis; int level; int visl,…

题意: 输入两个正整数N和K(2<=N<=200),代表城市的数量和道路的数量.接着输入起点城市的名称(所有城市的名字均用三个大写字母表示),接着输入N-1行每行包括一个城市的名字和到达该城市所能获得的快乐,接着输入M行每行包括一条道路的两端城市名称和道路的长度.输出从起点城市到目标城市"ROM"获得最大快乐且经历道路长度最短的道路条数和经历的道路长度和获得的快乐总数以及其中经过城市最少的那条路的到达每个城市所获得的平均快乐.下一行输出这条路的路径,城市名之间用"…

题目分析: 这题我在写的时候在PTA提交能过但是在牛客网就WA了一个点,先写一下思路留个坑 这题的简单来说就是需要找一条最短路->最开心->点最少(平均幸福指数自然就高了),由于本题给出的所有的城市的名称都是字符串(三个大写字母),所以我先将每个字符串计算一个hash值去代表每一个城市,接着就是用dijkstra算法去逐步判断各种情况,是直接用了一个dijkstra去计算,而没有说是先dijkstra算出最短距离再dfs所有的路径去找出最优解,因为我觉得这题局部最优就能反映到全局最优,也不知是…

预处理每个节点左子树有多少个点. 然后确定值得时候递归下去就可以了. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; struct Node { int…

根据中序遍历和前序遍历确定一棵二叉树,然后按“层次遍历”序列输出.输出规则:除根节点外,接下来每层的节点输出顺序是:先从左到右,再从右到左,交替输出 #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <string> #include <map> #define LEFT 0 #define RIGHT 1 u…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…