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 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
思路
模拟一下AVL树的过程。
四个旋转方法,左单旋,右单旋,左右双旋,右左双旋。具体看代码,函数名写得挺清楚的。
还需要一个高度函数,递归一下就得出来了。
最后再弄一个insert
函数,注意AVL树的特点,左右子树的高度差为2时必须发生平衡旋转。
code
#include <iostream>
using namespace std;
struct node{
int data;
node *left, *right;
node(int data){this->data = data, this->left = this->right = NULL;}
};
node* rotate_left(node* root){
node* temp = root->left;
root->left = temp->right;
temp->right = root;
return temp;
}
node* rotate_right(node* root){
node* temp = root->right;
root->right = temp->left;
temp->left = root;
return temp;
}
node* rotate_left_right(node* root){
root->right = rotate_left(root->right);
return rotate_right(root);
}
node* rotate_right_left(node* root){
root->left = rotate_right(root->left);
return rotate_left(root);
}
int getHeight(node* root){
if(root == NULL) return 0;
return max(getHeight(root->right), getHeight(root->left)) + 1;
}
node* insert(int data, node* root){
if(root == NULL) root = new node(data);
else if(data > root->data) {
root->right = insert(data, root->right);
if(getHeight(root->right) - getHeight(root->left) >= 2)
root = root->right->data > data ? rotate_left_right(root) : rotate_right(root);
}
else {
root->left = insert(data, root->left);
if(getHeight(root->left) - getHeight(root->right) >= 2)
root = root->left->data < data ? rotate_right_left(root) : rotate_left(root);
}
return root;
}
int main(){
int n = 0, temp = 0;
scanf("%d", &n);
node *root = NULL;
for(int i = 0; i < n; i++){
scanf("%d", &temp);
root = insert(temp, root);
}
printf("%d", root->data);
return 0;
}
PAT甲级:1066 Root of AVL Tree (25分)的更多相关文章
- 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 ...
- 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 ...
- PAT甲级1066. Root of AVL Tree
PAT甲级1066. Root of AVL Tree 题意: 构造AVL树,返回root点val. 思路: 了解AVL树的基本性质. AVL树 ac代码: C++ // pat1066.cpp : ...
- PAT 甲级 1066 Root of AVL Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805404939173888 An AVL tree is a self- ...
- PAT Advanced 1066 Root of AVL Tree (25) [平衡⼆叉树(AVL树)]
题目 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child ...
- 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 ...
- 【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 ...
- 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 ...
- PAT 1066. Root of AVL Tree (25)
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
随机推荐
- Centos8配置Nginx开机自启动
第一步:创建service文件,并编辑(可理解为开机时自动启动Nginx的脚本服务文件) vim /lib/systemd/system/nginx.service /lib 与 /usr/lib 里 ...
- java接口类
是什么:类似于java中的继承,但是继承只可以继承一个人父类,接口类可以继承多个 作用:解决java继承解决不了的问题 关键字:interface(定义) implements(使用) 注意事项:1. ...
- ES5中的类
之前小编对于类和类的基本特征(所谓的封装.继承.多态)理解一直不是很到位,同时在实际项目应用中也用的比较少,今天小编就结合ES5中的基本用法和大家聊聊,希望小伙伴会在这篇文章有属于自己的收获,并能够在 ...
- frp实现NAS外网访问
开始 入手了NAS之后就看是漫漫的折腾内网穿透的不归路.用的最多最简单的是下面三种 QC:群晖自带的,速度在100kb左右,能不能连上有时候都是运气,极其不稳定. DDNS:最推荐的方式.上下行都取决 ...
- 【NX二次开发】Block UI 目录
Block UI 目录 基本 标签/位图 切换开关 枚举 字符串 多行字符串 操作按钮 列表框 分割线 对象颜色选择器 RGB颜色选择器 绘图区 图层 ...
- 【NX二次开发】Block UI 指定方位
属性说明 属性 类型 描述 常规 BlockID String 控件ID Enable Logical 是否可操作 Group ...
- NX二次开发-通过3x3矩阵获取XYZ轴矢量
函数:UF_CSYS_ask_wcs() 函数说明:通过3x3矩阵获取XYZ轴矢量 用法: 1 #include <uf.h> 2 #include <uf_mtx.h> 3 ...
- DB2 SQL0805N解决和思考
一.报错现象 这是一个在使用 DB2数据库过程中比较常见的错误, 报错信息如下 Exception stack trace: com.ibm.db2.jcc.am.SqlException: DB2 ...
- VBS脚本编程(4)——流程控制语句
分支结构--If .. Then .. Else .. 根据表达式的值有条件地执行一组语句. If condition Then statements [Else elsestatements ] 或 ...
- Vue(6)v-on指令的使用
v-on 监听事件 可以用 v-on 指令监听 DOM 事件,并在触发时运行一些 JavaScript 代码.事件代码可以直接放到v-on后面,也可以写成一个函数.示例代码如下: <div id ...