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 (≤) which is the total number of keys to be inserted. Then Ndistinct 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;
int height;
Node* lchild;
Node* rchild;
}*root; int getHeight(Node* root);
void updateHeight(Node* root);
int getBalanceFactor(Node* root);
Node* NewNode(int v);
void Insert(Node* &root, int v);
void L(Node* &root);
void R(Node* &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 ;
}
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);
}
}
}
} Node* NewNode(int v){
Node* node = new Node;
node->v = v;
node->lchild = node->rchild = NULL;
node->height = ;
return node;
} void updateHeight(Node* root){
root->height = max(getHeight(root->lchild),getHeight(root->rchild))+;
} int getHeight(Node* root){
if(root == NULL) return ;
return root->height;
} int getBalanceFactor(Node* root){
return getHeight(root->lchild) - getHeight(root->rchild);
} void L(Node* &root){
Node* temp = root->rchild;
root->rchild = temp->lchild;
temp->lchild = root;
updateHeight(root);
updateHeight(temp);
root = temp;
} void R(Node* &root){
Node* temp = root->lchild;
root->lchild = temp->rchild;
temp->rchild = root;
updateHeight(root);
updateHeight(temp);
root = temp;
}

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. Mockplus组件样式库一键解决风格复用

    在Mockplus3.3版本中,新增了组件样式库,可以快速复用组件风格,同时可以将组件风格保存到库中. 官网地址:https://www.mockplus.cn 1. 保存样式 选中组件,设置好该组件 ...

  2. 签名Android应用程序

    Android要求对作为产品发布的应用进行签名(包名相同的化,后安装的应用会覆盖前面安装的应用) 签名作用:1.确定发布者的身份.2.确保应用的完整性. 注意:在应用的开发.调试阶段,Eclipse的 ...

  3. Android 4.0关于开机启动服务

    针对使用App应用管理强制停止的App,重启系统后不能收到开机启动, 需要运行一次后,在下次再启动时,才可以正确收到.

  4. 51nod1459迷宫问题—(迪杰斯特拉)

    1459 迷宫游戏  基准时间限制:1 秒 空间限制:131072 KB 分值: 0 难度:基础题  收藏  关注 你来到一个迷宫前.该迷宫由若干个房间组成,每个房间都有一个得分,第一次进入这个房间, ...

  5. Low-level Thinking in High-level Shading Languages

    因为要反汇编shader代码,所以google了数学函数_sat的知识,发现了一些高级着色语言的优化相关的问题.Low-level Thinking in High-level Shading Lan ...

  6. How to generate HMAC-SHA1 in C#?

    using (HMACSHA1 m = new HMACSHA1(Encoding.UTF8.GetBytes("AKIDZfbOA78asKUYBcXFrJD0a1ICvR98JM&quo ...

  7. [Elixir005] 查看指定数据的详细信息 i helper

    elixir在1.2后增加了一个新的特性i helper. 在iex shell中使用i可以查看任意数据的数据类型和详细描述 #查看变量描述 iex(1)> i {:test, "Th ...

  8. C# Session操作

    Session.Abandon();//清除全部Session//清除某个SessionSession["UserName"] = null;Session.Remove(&quo ...

  9. C#设计模式(23种模式)

    https://www.cnblogs.com/abcdwxc/archive/2007/10/30/942834.html

  10. MySQL多线程备份工具mydumper 之 RDS外部实例迁移平台

    此文已由作者温正湖授权网易云社区发布. 欢迎访问网易云社区,了解更多网易技术产品运营经验. 1.Format_description_event问题: BINLOG ' kTXkUxMKAAAALQA ...