PAT_A1123#Is It a Complete AVL Tree
Source:
Description:
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, orNO
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
Keys:
- 二叉树的建立
- 二叉树的遍历
- 完全二叉树(Complete Binary Tree)
- 平衡二叉树(Self-balancing Binary Search Tree,AVL tree)
Attention:
- 判断完全二叉树,while的固定写法
- Rotation中,先Update root,再Update temp,否则会影响结果,注意
Code:
/*
Data: 2019-06-24 15:36:45
Problem: PAT_A1123#Is It a Complete AVL Tree
AC: 35:46 题目大意:
由插入序列构造一棵AVL树,输出层次遍历并判断是否为一棵完全二叉树 基本思路:
构造平衡二叉树,
中序遍历并判断是否为完全二叉树
*/
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
int data;
int height;
node *lchild, *rchild;
}; int GetHeight(node *root)
{
if(root == NULL)
return ;
else
return root->height;
} int GetBalanceFactor(node *root)
{
return GetHeight(root->lchild) - GetHeight(root->rchild);
} void UpdataHeight(node *&root)
{
root->height = max(GetHeight(root->lchild),GetHeight(root->rchild))+;
} void LeftRotation(node *&root)
{
node *temp = root->rchild;
root->rchild = temp->lchild;
temp->lchild = root;
UpdataHeight(root);
UpdataHeight(temp);
root = temp;
} void RightRotation(node *&root)
{
node *temp = root->lchild;
root->lchild = temp->rchild;
temp->rchild = root;
UpdataHeight(root);
UpdataHeight(temp);
root = temp;
} void Insert(node *&root, int x)
{
if(root == NULL)
{
root = new node;
root->data = x;
root->height=;
root->lchild = root->rchild = NULL;
}
else if(x < root->data)
{
Insert(root->lchild, x);
UpdataHeight(root);
if(GetBalanceFactor(root) == )
{
if(GetBalanceFactor(root->lchild) == )
RightRotation(root);
else
{
LeftRotation(root->lchild);
RightRotation(root);
}
}
}
else
{
Insert(root->rchild, x);
UpdataHeight(root);
if(GetBalanceFactor(root) == -)
{
if(GetBalanceFactor(root->rchild) == -)
LeftRotation(root);
else
{
RightRotation(root->rchild);
LeftRotation(root);
}
}
}
} int IsComplete(node *root, int n)
{
queue<node*> q;
q.push(root);
int cnt=, ans=;
while(!q.empty())
{
root = q.front();
q.pop();
if(root)
{
printf("%d%c", root->data,++cnt==n?'\n':' ');
q.push(root->lchild);
q.push(root->rchild);
}
else
{
if(cnt==n)
break;
else
{
ans=;
while(!q.empty())
{
root = q.front();
if(root) break;
else q.pop();
}
}
}
}
return ans;
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE int n,x;
node *root = NULL;
scanf("%d", &n);
for(int i=; i<n; i++)
{
scanf("%d", &x);
Insert(root, x);
}
if(IsComplete(root, n))
printf("YES");
else
printf("NO"); return ;
}
PAT_A1123#Is It a Complete AVL Tree的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- A1123. Is It a Complete AVL Tree
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
- 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 ...
- 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 ...
- 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 ...
- PAT 1123 Is It a Complete AVL Tree
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
- PAT甲级——A1123 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 ...
随机推荐
- CODEVS——T 1049 棋盘染色
http://codevs.cn/problem/1049/ 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 黄金 Gold 题解 查看运行结果 题目描述 Descr ...
- Spring MVC-控制器(Controller)-属性方法名称解析器(Properties Method Name Resolver )示例(转载实践)
以下内容翻译自:https://www.tutorialspoint.com/springmvc/springmvc_propertiesmethodnameresolver.htm 说明:示例基于S ...
- HDU 5288
//枚举因子,查找和i最近的左右是i因子的点即可. #include <iostream> #include <cstdio> #include <algorithm&g ...
- springmvc学习笔记(10)-springmvc注解开发之商品改动功能
springmvc学习笔记(10)-springmvc注解开发之商品改动功能 标签: springmvc springmvc学习笔记10-springmvc注解开发之商品改动功能 需求 开发mappe ...
- Python常用模块【sys】
sys.argv 参数 「argv」是「argument variable」参数变量的简写形式.一般在命令行调用的时候由系统传递给程序.这个变量其实是一个List列表,argv[0] 一般是“被 ...
- Web实际应用中的编码问题
一. JSP页面有关编码的介绍 ---->>假设不做不论什么设置,页面默认ISO-8859-1编码(Western European). ---->><%@ page c ...
- UESTC--1253--阿里巴巴和n个大盗 (博弈)
阿里巴巴和n个大盗 Time Limit: 1000MS Memory Limit: 65535KB 64bit IO Format: %lld & %llu Submit St ...
- B1003 物流运输(最短路 + dp)
这个dp其实不是那么难,状态其实很好想,但是细节有少许偏差. 当时我并没有想到最短路要在dp之外写,后来看题解之后发现要预处理出来每段时间1~M的最短路,然后直接dp. 题目: Description ...
- PCB MS SQL 将字符串分割为表变量(表值函数)
Create FUNCTION [dbo].[SplitTable]( @s varchar(max), --待分拆的字符串 ) --数据分隔符 ),), col varchar(max)) --建立 ...
- 洛谷P1040 加分二叉树(区间dp)
P1040 加分二叉树 题目描述 设一个n个节点的二叉树tree的中序遍历为(1,2,3,…,n),其中数字1,2,3,…,n为节点编号.每个节点都有一个分数(均为正整数),记第i个节点的分数为di, ...