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 illustrate the rotation rules.



Now given a sequence of insertions, you are supposed to output the level-order traversal sequence of the resulting AVL tree, and to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤ 20). Then N distinct integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, insert the keys one by one into an initially empty AVL tree. Then first print in a line the level-order traversal sequence of the resulting AVL tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line. Then in the next line, print YES if the tree is complete, or NO if not.

Sample Input 1:

5

88 70 61 63 65

Sample Output 1:

70 63 88 61 65

YES

Sample Input 2:

8

88 70 61 96 120 90 65 68

Sample Output 2:

88 65 96 61 70 90 120 68

NO

#include<iostream>
#include<math.h>
#include<queue>
using namespace std;
struct node{
int value, depth;
node* l=NULL;
node* r=NULL;
node(int v): value(v), depth(0), l(NULL), r(NULL){
}
};
int getheight(node* root){
return root==NULL?0:max(getheight(root->l), getheight(root->r))+1;
} node* RotationLL(node* root){
node* temp=root->l;
root->l=temp->r;
temp->r=root;
temp->depth=getheight(temp);
root->depth=getheight(root);
return temp;
} node* RotationRR(node* root){
node* temp=root->r;
root->r=temp->l;
temp->l=root;
temp->depth=getheight(temp);
root->depth=getheight(root);
return temp;
} node* RotationLR(node* root){
root->l=RotationRR(root->l);
return RotationLL(root);
} node* RotationRL(node* root){
root->r=RotationLL(root->r);
return RotationRR(root);
} node* insert(node* root, int val){
if(root==NULL){
root=new node(val);
return root;
}else if(val<root->value){
root->l=insert(root->l, val);
if(getheight(root->l)-getheight(root->r)==2)
if(val<root->l->value)
root=RotationLL(root);
else
root=RotationLR(root); }else{
root->r=insert(root->r, val);
if(getheight(root->l)-getheight(root->r)==-2)
if(val<root->r->value)
root=RotationRL(root);
else
root=RotationRR(root);
}
root->depth=getheight(root);
return root;
}
int main(){
int n, flag=0, ans=0, first=0;
cin>>n;
node* root=NULL;
for(int i=0; i<n; i++){
int val;
cin>>val;
root=insert(root, val);
}
queue<node*> q;
q.push(root);
while(!q.empty()){
node* temp=q.front();
first++==0?cout<<temp->value:cout<<" "<<temp->value;
q.pop();
if(temp->l!=NULL){
q.push(temp->l);
flag==1?ans=1:ans=ans;
}
else
flag=1;
if(temp->r!=NULL){
q.push(temp->r);
flag==1?ans=1:ans=ans;
}
else
flag=1;
}
cout<<endl;
ans==1?cout<<"NO"<<endl:cout<<"YES"<<endl;
return 0;
}

PAT 1123 Is It a Complete AVL Tree的更多相关文章

  1. PAT 1123. Is It a Complete AVL Tree (30)

    AVL树的插入,旋转. #include<map> #include<set> #include<ctime> #include<cmath> #inc ...

  2. PAT甲级1123. Is It a Complete AVL Tree

    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 ...

  3. 1123 Is It a Complete AVL Tree

    1123 Is It a Complete AVL Tree(30 分) An AVL tree is a self-balancing binary search tree. In an AVL t ...

  4. PAT甲级——1123 Is It a Complete AVL Tree (完全AVL树的判断)

    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 sel ...

  5. PAT Advanced 1123 Is It a Complete AVL Tree (30) [AVL树]

    题目 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child ...

  6. 1123. Is It a Complete AVL Tree (30)

    An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...

  7. PAT A1123 Is It a Complete AVL Tree (30 分)——AVL平衡二叉树,完全二叉树

    An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...

  8. 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 sub ...

  9. PAT甲级1123 Is It a Complete AVL Tree【AVL树】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805351302414336 题意: 给定n个树,依次插入一棵AVL ...

随机推荐

  1. 【145】◀▶ .NET Framework类库索引

    C#编程基础: A1 ………… 基础A2 ………… using 关键字A3 ………… as 关键字A4 ………… is 关键字A5 ………… switch 关键字A6 ………… return 语句关键 ...

  2. clc和clear命令的使用

    clc命令是用来清除命令窗口的内容,这点不用多说.不管开启多少个应用程序,命令窗口只有一个,所以clc无论是在脚本m文件或者函数m文件调用时,clc命令都会清除命令窗口的内容.clear命令可以用来清 ...

  3. 如何过滤 adb logcat 输出(转载)

    转自:http://www.cnblogs.com/imouto/archive/2012/12/11/filtering-adb-logcat-output.html 简介: 本文介绍如何在 she ...

  4. Linux 系统管理命令 - uptime - 显示系统的运行时间及负载

    命令详解 重要星级: ★★★☆☆ 功能说明: uptime 命令可以输出当前系统时间.系统开机到现在的运行时间.目前有多少用户在线和系统平均负载等信息 语法格式: uptime 说明: 直接执行 up ...

  5. bzoj 3743: [Coci2015]Kamp【树形dp】

    两遍dfs一遍向下,一边向上,分别记录子树内人数a,当前点到所有点的距离b,最大值c和次大值d,最大值子树p 然后答案是2b-c #include<iostream> #include&l ...

  6. robotframework - selenium 分层思路

    前言: 对于每一条用例来说,调用“百度搜索”关键字,输入搜索内容,输入预期结果即可.不同关心用例是如何执行的.如果百度输入框的定位发生了变化,只用去修改“百度搜索”关键字即可,不用对每一条用例做任何修 ...

  7. May Challenge 2019 Division 2 水题讲解

    Reduce to One 这题其实蛮水的? 题意就是说: 给定一个 1~n 的序列,每次挑两个数 x y 合并,合并值为 \(x+y+xy\) ,然后求不断合并最后剩下的一个的最大值 随便搞搞发现答 ...

  8. [转]C语言常见错误总结1

    指针与数组的对比c程序中,指针和数组在不少地方可以相互替换着用,让人产生一种错觉,以为两者是等价的 数组要么在静态存储区被创建(如全局数组),要么在栈上被创建.数组名对应着(而不是指向)一块内存,其地 ...

  9. 如何在Ubuntu上安装Wine 2.6

    Wine(Wine不是模拟器)是一种开源兼容层软件应用程序,可以让Linux和Unix用户通过Winelib软件库在他们的系统上运行Windows软件. sudo add-apt-repository ...

  10. YumRepo Error: All mirror URLs are not using ftp, http[s] or file

    有台机器使用Yum的时候,报错如下: YumRepo Error: All mirror URLs are not using ftp, http[s] or file. Eg. $releaseve ...