PAT 甲级 1064 Complete Binary Search Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805407749357568
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
代码:
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + 10;
int N;
int a[maxn];
vector<int> level; void levelorder(int st, int en, int index) {
if(st > en) return ;
int n = en - st + 1;
int l = log(n + 1) / log(2);
int leave = n - (pow(2, l) - 1);
int root = st + (pow(2, l - 1) - 1) + min((int)pow(2, l - 1), leave);
level[index] = a[root];
levelorder(st, root - 1, 2 * index + 1);
levelorder(root + 1, en, 2 * index + 2);
} int main() {
scanf("%d", &N);
level.resize(N);
for(int i = 0; i < N; i ++)
scanf("%d", &a[i]); sort(a, a + N);
levelorder(0, N - 1, 0);
for(int i = 0; i < N; i ++) {
printf("%d", level[i]);
printf("%s", i != N - 1 ? " " : "\n");
}
return 0;
}
https://www.liuchuo.net/archives/2161
还是自己太菜了 打扰了!
FHFHFH
PAT 甲级 1064 Complete Binary Search Tree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 ) (数据结构)
1064 Complete Binary Search Tree (30 分) A Binary Search Tree (BST) is recursively defined as a binar ...
- PAT甲级——A1064 Complete Binary Search Tree
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following propertie ...
- 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 ...
- 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 1064 Complete Binary Search Tree[二叉树][难]
1064 Complete Binary Search Tree (30)(30 分) A Binary Search Tree (BST) is recursively defined as a b ...
- 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 Tr ...
随机推荐
- kali国内更新源
在/tec/apt/sources.list加入以下内容 #中科大更新源 deb https://mirrors.ustc.edu.cn/kali kali-rolling main non-free ...
- 【转】枚举enum学习小记
原帖: http://hi.baidu.com/yuleishou/item/caacae872190031ec216272f 表示在vs2008下实验了一下,有些东西和原帖的还是不一样的,都贴在这里 ...
- 【转载】MFC的Main函数跑哪去了
原文:http://blog.csdn.net/weiwenhp/article/details/8455471 习惯的思维 用习惯了C的人要看一个程序时首先会想到找到那个main函数在哪,然后再顺着 ...
- CF1039D You Are Given a Tree 根号分治,贪心
CF1039D You Are Given a Tree LG传送门 根号分治好题. 这题可以整体二分,但我太菜了,不会. 根号分治怎么考虑呢?先想想\(n^2\)暴力吧.对于每一个要求的\(k\), ...
- SpringCloud-初识微服务(一)
前言 本篇文章简单介绍一下什么是微服务.微服务的优点.SpringCloud的微服务架构核心组件选型等: 一.什么是微服务? 微服务的提出者Martin Fowler是这样描述微服务的(原文:http ...
- 2_C语言中的数据类型 (四)整数与无符号数
1.1 sizeof关键字 sizeof是c语言关键字,功能是求指定数据类型在内存中的大小,单位:字节 sizeof与size_t类型 1.1 int类型 1.1.1 ...
- IDEA 出现 updating indices 卡进度条问题的解决方案并加快索引速度
缺点: 这样的话,前端的接口(也就是字符串)就搜索不到了. C:\Users\Administrator\.IntelliJIdea2017.3\system 删除里面的caches文件夹(这里的 ...
- java.util.Arrays.asList 的小问题
JDK 1.4对java.util.Arrays.asList的定义,函数参数是Object[].所以,在1.4中asList()并不支持基本类型的数组作参数. JDK 1.5中,java.util. ...
- leetcode刷题笔记191 位1的个数
题目描述: 编写一个函数,输入是一个无符号整数,返回其二进制表达式中数字位数为 ‘1’ 的个数(也被称为汉明重量). 示例: 输入: 输出: 解释: 32位整数 的二进制表示为 . 题目分析: 判断3 ...
- k8s踩坑记第2篇--3个IP折磨人的故事
例子来源于<Kubernetes实践指南>一书.问题依然没有解决,求助大神. 测试环境 Centos 7.0 docker 1.13.1 kubectl v1.5.2 etcd 3.2.1 ...