1064. Complete Binary Search Tree (30)【二叉树】——PAT (Advanced Level) Practise
题目信息
1064. Complete Binary Search Tree (30)
时间限制100 ms
内存限制65536 kB
代码长度限制16000 B
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
- The left subtree of a node contains only nodes with keys less than the node’s key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
- Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.
Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding complete binary search 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.
Sample Input:
10
1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
解题思路
模拟建树推出层序
AC代码
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int n, a[1005];
vector<int> level[22];
void step(int loc, int len, int lv){
if (len <= 0) return;
int t = 1;
while (t*2 <= len) t *= 2;
--t;
int cd = ((len - (len - t)) + 1) / 2 + len - t;
if (cd <= t + 1){
level[lv].push_back(a[loc + cd - 1]);
step(loc, cd - 1, lv + 1);
step(loc + cd, len - cd, lv + 1);
}else{
level[lv].push_back(a[loc + t]);
step(loc, t, lv + 1);
step(loc + t + 1, len - t - 1, lv + 1);
}
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; ++i){
scanf("%d", a+i);
}
sort(a, a + n);
step(0, n, 0);
printf("%d", level[0][0]);
for (int i = 1; i <= 13; ++i){
for (int j = 0; j < level[i].size(); ++j){
printf(" %d", level[i][j]);
}
}
printf("\n");
}
个人游戏推广: apkName=com.xianyun.yf" target="_blank" align="left"> apkName=com.xianyun.yf" target="_blank" align="left">《10云方》与方块来次消除大战!
1064. Complete Binary Search Tree (30)【二叉树】——PAT (Advanced Level) Practise的更多相关文章
- pat 甲级 1064. Complete Binary Search Tree (30)
1064. Complete Binary Search Tree (30) 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 CHE ...
- PAT 甲级 1064 Complete Binary Search Tree (30 分)(不会做,重点复习,模拟中序遍历)
1064 Complete Binary Search Tree (30 分) A Binary Search Tree (BST) is recursively defined as a bin ...
- PAT甲级:1064 Complete Binary Search Tree (30分)
PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as ...
- PAT题库-1064. Complete Binary Search Tree (30)
1064. Complete Binary Search Tree (30) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...
- PAT Advanced 1064 Complete Binary Search Tree (30) [⼆叉查找树BST]
题目 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following proper ...
- 1064 Complete Binary Search Tree (30分)(已知中序输出层序遍历)
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following propertie ...
- PAT甲题题解-1064. Complete Binary Search Tree (30)-中序和层次遍历,水
由于是满二叉树,用数组既可以表示父节点是i,则左孩子是2*i,右孩子是2*i+1另外根据二分搜索树的性质,中序遍历恰好是从小到大排序因此先中序遍历填充节点对应的值,然后再层次遍历输出即可. 又是一道遍 ...
- 【PAT甲级】1064 Complete Binary Search Tree (30 分)
题意:输入一个正整数N(<=1000),接着输入N个非负整数(<=2000),输出完全二叉树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TI ...
- PAT (Advanced Level) 1064. Complete Binary Search Tree (30)
因为是要构造完全二叉树,所以树的形状已经确定了. 因此只要递归确定每个节点是多少即可. #include<cstdio> #include<cstring> #include& ...
随机推荐
- Hive扩展功能(九)--Hive的行级更新操作(Update)
软件环境: linux系统: CentOS6.7 Hadoop版本: 2.6.5 zookeeper版本: 3.4.8 主机配置: 一共m1, m2, m3这三部机, 每部主机的用户名都为centos ...
- unbuntu系统:python2.7安装pyspark
以前在进行搜索引擎rank-svm排序模型训练时,直接使用python读取的HDFS日志文件.统计计算等预处理操作再进行svm模型,最终产生出训练模型.现在回想一下,数据预处理这一块完全可以使用spa ...
- 北大ACM(POJ1019-Number Sequence)
Question:http://poj.org/problem?id=1019 问题点:打表. Memory: 392K Time: 16MS Language: C++ Result: Accept ...
- MySql学习笔记(三) —— 聚集函数的使用
1.AVG() 求平均数 select avg(prod_price) as avg_price from products; --返回商品价格的平均值 ; --返回生产商id为1003的商品价格平均 ...
- 微信小程序 客服自动回复图片
产品需求是,在客服对话框里,发送特定的文字,回复我们的二维码: 小城程开发完成后,这个自动回复图片的功能就摆在了眼前.刚开始我们想到的是:在线客服功能的设置里设置好自动回复的图片,但是目前设置不支持自 ...
- vue基础---实例
(1)数据和方法 ①响应式双向绑定 当一个 Vue 实例被创建时,它向 Vue 的响应式系统中加入了其 data 对象中能找到的所有的属性.当这些属性的值发生改变时,视图将会产生“响应”,即匹配更新为 ...
- UVA - 12661 Funny Car Racing (Dijkstra算法)
题目: 思路: 把时间当做距离利用Dijkstra算法来做这个题. 前提:该结点e.c<=e.a,k = d[v]%(e.a+e.b); 当车在这个点的1处时,如果在第一个a这段时间内能够通过且 ...
- TestNG常用注解
原文链接:https://www.yiibai.com/testng/basic-annotations.html 以下是TestNG支持的注释列表: 注解 描述 @BeforeSuite 在该 ...
- POJ2528 Uva10587 Mayor's posters
The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign h ...
- [BZOJ1045][HAOI2008]糖果传递(数学分析)
题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1045 分析:均分纸牌的环状版本. 先看线性的版本: 设f[i]表示第I位从第i+1位得 ...