PAT-1066(Root of AVL Tree)Java语言实现
Root of AVL Tree
PAT-1066
- 这是关于AVL即二叉平衡查找树的基本操作,包括旋转和插入
- 这里的数据结构主要在原来的基础上加上节点的高度信息。
import java.util.*;
/**
* @Author WaleGarrett
* @Date 2020/9/5 10:41
*/
public class PAT_1066 {
public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
int n=scanner.nextInt();
AVLNode root=null;
while(n!=0){
int value=scanner.nextInt();
root=insert(root,value);
// printTree(root);
n--;
}
System.out.println(root.value);
}
static void printTree(AVLNode root){
List<AVLNode> list=new ArrayList<>();
list.add(root);
while(list.size()!=0){
AVLNode temp=list.remove(0);
System.out.print(temp.value+" ");
if(temp.left!=null)
list.add(temp.left);
if(temp.right!=null)
list.add(temp.right);
}
System.out.println();
}
/**
* 顺时针旋转
* @param root
* @return
*/
public static AVLNode rightRotate(AVLNode root){
AVLNode temp=root.left;
root.left=temp.right;
temp.right=root;
temp.updateHeight();
root.updateHeight();
return temp;
}
/**
* 逆时针旋转
* @param root
* @return
*/
public static AVLNode leftRotate(AVLNode root){
AVLNode temp=root.right;
root.right=temp.left;
temp.left=root;
temp.updateHeight();
root.updateHeight();
return temp;
}
/**
* 向平衡二叉排序树里插入一个节点
* @param value
*/
public static AVLNode insert(AVLNode root,int value){
if(root==null){
root=new AVLNode(null,null,value,1);
return root;
}
if(value<root.value){
root.left=insert(root.left,value);//插入根节点的左子树中
root.updateHeight();
if(root.getBalanceFactor()>1){//当前节点不平衡
if(root.left.getBalanceFactor()>0){//LL插入
root=rightRotate(root);
}else if(root.left.getBalanceFactor()<0){//LR插入
root.left=leftRotate(root.left);
root=rightRotate(root);
}
}
}else if(value>root.value){
root.right=insert(root.right,value);
root.updateHeight();
if(root.getBalanceFactor()<-1){//当前节点不平衡
if(root.right.getBalanceFactor()<0){//RR插入
root=leftRotate(root);
}else if(root.right.getBalanceFactor()>0){//RL插入
root.right=rightRotate(root.right);
root=leftRotate(root);
}
}
}
return root;
}
}
class AVLNode{
AVLNode left;
AVLNode right;
int value;
private int height;//该结点的高度
public AVLNode(){
left=right=null;
value=-1;
height=0;
}
public AVLNode(AVLNode left,AVLNode right,int value,int height){
this.value=value;
this.left=left;
this.right=right;
this.height=height;
}
public int getHeight() {
return height;
}
public int getBalanceFactor(){
int leftHeight,rightHeight;
if(left==null)
leftHeight=0;
else leftHeight=left.getHeight();
if(right==null)
rightHeight=0;
else rightHeight=right.getHeight();
return leftHeight-rightHeight;
}
void updateHeight(){
int leftHeight,rightHeight;
if(left==null)
leftHeight=0;
else leftHeight=left.getHeight();
if(right==null)
rightHeight=0;
else rightHeight=right.getHeight();
height=Math.max(leftHeight,rightHeight)+1;
}
}
PAT-1066(Root of AVL Tree)Java语言实现的更多相关文章
- PAT 1066 Root of AVL Tree[AVL树][难]
1066 Root of AVL Tree (25)(25 分) An AVL tree is a self-balancing binary search tree. In 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 ...
- PAT 1066 Root of AVL Tree
#include <cstdio> #include <cstdlib> class Node { public: Node* L; Node* R; int height; ...
- 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 (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 ...
- 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 (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 ...
- PTA (Advanced Level) 1066 Root of AVL Tree
Root of AVL Tree An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of ...
- 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 ...
随机推荐
- hdu 6860 Fluctuation Limit 双向贪心
题意: 给你n个区间[li,ri],和一个整数k,你从每一个区间内选出来一个数,把从第i个区间内选出来数放在第i个位置,这样会构成一个长度为n的序列,你需要保证序列中任意两个相邻的数之差的绝对值要小于 ...
- Codeforces Round #645 (Div. 2) D. The Best Vacation (贪心,二分)
题意:一年有\(n\)个月,每月有\(d_{i}\)天,找出连续的\(x\)天,使得这\(x\)天的日期总和最大,任意一年都能选. 题解:首先要先贪心,得到:连续的\(x\)天的最后一天一定是某个月的 ...
- CF1463-C. Busy Robot
题意: 你有一个机器人,这个机器人在一维坐标轴上移动.你可以给这个机器人下达指令,指令的形式为 \(t_i, x_i\) ,意味着机器人在第\(t_i\)秒的时候获得一条指令,此时这个机器人以\(1/ ...
- HTTP常见状态码(200、301、302、404、500、502)详解
概述 运维工作中,在应用部署的时候,通常遇到各种HTTP的状态码,我们比较常见的如:200.301.302.404.500.502 等,有必要整理一份常见状态码的文档,加深印象,方便回顾. ...
- codeforces 1042D - Petya and Array【树状数组+离散化】
题目:戳这里 题意:有n个数,问有多少个区间满足[L,R]内的和小于t. 解题思路: [L,R]内的和小于t等价于sum[R]-sum[L-1]<t,将sum[L-1]左移,可以看出R与L的关系 ...
- python sort() sorted()的区别
sorted不保存 sort保存:
- JS编程练习:将目标节点内部的子节点逆序
将目标节点内部的子节点逆序 1 <body> 2 <div> 3 <p></p> 4 <span></span> 5 <e ...
- 深入剖析JavaScript中的对象与原始值之间的转换机制
我们都知道原始值之间是可以互相转换的,但是如果对象转原始值呢? 所有的对象在布尔上下文(context)中均为 true .所以对于对象,不存在 to-boolean 转换, 只有字符串和数值转换. ...
- webpack 5 模块联合
webpack 5 模块联合 webpack 5 https://webpack.docschina.org/concepts/module-federation/ https://github.co ...
- GitHub SSH key
GitHub SSH key https://help.github.com/en/github/authenticating-to-github steps HTTPS clone !== SSH ...