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 tell the root of the resulting AVL tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤20) which is the total number of keys to be inserted. 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, print the root of the resulting AVL tree in one line.

Sample Input 1:

5
88 70 61 96 120

Sample Output 1:

70

Sample Input 2:

7
88 70 61 96 120 90 65

Sample Output 2:

88

#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
int v,height;
node* lchild,*rchild;
}*root; node* newNode(int v){
node* Node = new node;
Node->v = v;
Node->height = ;
Node->lchild = Node->rchild = NULL;
return Node;
} int getHeight(node* root){
if(root == NULL) return ;
return root->height;
} void updateHeight(node* root){
root->height = max(getHeight(root->lchild),getHeight(root->rchild)) + ;
} int getBalanceFactor(node* root){
return getHeight(root->lchild) - getHeight(root->rchild);
} void R(node* &root){
node* temp = root->lchild;
root->lchild = temp->rchild;
temp->rchild = root;
updateHeight(root);
updateHeight(temp);
root = temp;
} void L(node* &root){
node* temp = root->rchild;
root->rchild = temp->lchild;
temp->lchild = root;
updateHeight(root);
updateHeight(temp);
root = temp;
} void insert(node* &root,int v){
if(root == NULL){
root = newNode(v);
return;
}
if(root->v > v){
insert(root->lchild,v);
updateHeight(root);
if(getBalanceFactor(root) == ){
if(getBalanceFactor(root->lchild) == ){
R(root);
}else if(getBalanceFactor(root->lchild) == -){
L(root->lchild);
R(root);
}
}
}else{
insert(root->rchild,v);
updateHeight(root);
if(getBalanceFactor(root) == -){
if(getBalanceFactor(root->rchild) == -){
L(root);
}else if(getBalanceFactor(root->rchild) == ){
R(root->rchild);
L(root);
}
}
}
} int main(){
int n,v;
scanf("%d",&n);
for(int i = ; i < n; i++){
scanf("%d",&v);
insert(root,v);
}
printf("%d",root->v);
return ;
}

04-树5 Root of AVL Tree (25 分)的更多相关文章

  1. PTA 04-树5 Root of AVL Tree (25分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/668 5-6 Root of AVL Tree   (25分) An AVL tree ...

  2. PAT甲级:1066 Root of AVL Tree (25分)

    PAT甲级:1066 Root of AVL Tree (25分) 题干 An AVL tree is a self-balancing binary search tree. In an AVL t ...

  3. PAT 甲级 1066 Root of AVL Tree (25 分)(快速掌握平衡二叉树的旋转,内含代码和注解)***

    1066 Root of AVL Tree (25 分)   An AVL tree is a self-balancing binary search tree. In an AVL tree, t ...

  4. 1066 Root of AVL Tree (25分)(AVL树的实现)

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

  5. 04-树5 Root of AVL Tree (25 分)

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

  6. 【PAT甲级】1066 Root of AVL Tree (25 分)(AVL树建树模板)

    题意: 输入一个正整数N(<=20),接着输入N个结点的值,依次插入一颗AVL树,输出最终根结点的值. AAAAAccepted code: #define HAVE_STRUCT_TIMESP ...

  7. 04-树4. Root of AVL Tree (25)

    04-树4. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue An A ...

  8. pat04-树4. Root of AVL Tree (25)

    04-树4. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue An A ...

  9. pat 甲级 1066. Root of AVL Tree (25)

    1066. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An A ...

  10. pat1066. Root of AVL Tree (25)

    1066. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An A ...

随机推荐

  1. ubuntu 14 编译ARM g2o-20160424

    1. 安装eigen sudo apt-get install libeigen3-dev sudo apt-get install libsuitesparse-dev sudo apt-get i ...

  2. Luogu 4251 [SCOI2015]小凸玩矩阵

    BZOJ 4443 二分答案 + 二分图匹配 外层二分一个最小值,然后检验是否能选出$n - k + 1$个不小于当前二分出的$mid$的数.对于每一个$a_{i, j} \geq mid$,从$i$ ...

  3. keys()

    keys():返回一个数组,里面是符合匹配模式的键列表 $redis = new Redis(); $redis->connect('127.0.0.1', 6379); $pattern = ...

  4. Visual Studio 2010调试本地 IIS 站点

    点击vs的Debug-Attach to Process选中 w3wp.exe,然后点击Attach, vs便进入debug模式.

  5. CentOS7下源码包方式安装rabbitmq

    1.先安装erlang http://www.cnblogs.com/justphp/p/6093880.html 2.下载rabbitmq rpm包: wget http://www.rabbitm ...

  6. 利用Thread.stop完成方法执行超时中断

    示例代码可以从github上获取 https://github.com/git-simm/simm-framework.git 接上篇博客<FutureTask子线程取消执行的状态判断> ...

  7. javascript总结2: Date对象

    1 Date 对象 Date 对象用于处理日期与时间. Date()的方法很多,这里只总结工作必备的方法! 2 常用方法 创建个 Date 对象:const mydate=new Date(); &l ...

  8. 换零钞——第九届蓝桥杯C语言B组(国赛)第一题

    原创 标题:换零钞 x星球的钞票的面额只有:100元,5元,2元,1元,共4种.小明去x星旅游,他手里只有2张100元的x星币,太不方便,恰好路过x星银行就去换零钱.小明有点强迫症,他坚持要求200元 ...

  9. 装饰(Decorator)模式

    一. 装饰(Decorator)模式 装饰(Decorator)模式又名包装(Wrapper)模式[GOF95].装饰模式以对客户端透明的方式扩展对象的功能,是继承关系的一个替代方案. 二. 装饰模式 ...

  10. vue记住密码功能

    话不多说,直接上代码. html部分: <el-form :model="ruleForm2" :rules="rules2" ref="rul ...