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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 【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 ...

  9. 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

  1. 部署bugzilla(bugzilla+apache+mysql+linux)

    工作原因,需要部署bugzilla.在此,容我新造个轮子.官方轮子:https://bugzilla.readthedocs.org/en/latest/installing/quick-start. ...

  2. HDU_3732_(多重背包)

    Ahui Writes Word Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  3. HDU_3172_带权并查集

    Virtual Friends Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)T ...

  4. UIPageViewController 翻页、新手引导--UIScrollView:pagingEnabled

    UIPageViewController 翻页.新手引导--UIScrollView:pagingEnabled

  5. eclipse 新建 maven 项目 + 消除错误

    安装eclips以及maven自行去了解,这里不讲解 一.新建一个 maven 项目. 二.下一步选择项目架构 三.填写相关信息 group id: 一般都是 com点 项目名 aftriact.id ...

  6. JavaScript 中实现 sleep

    来自推特上 Windows 故障分析的笑话 图片来源:me.me 推上看到的笑话,Windows 故障分析的实现. 然后想起来 JavaScript 中如何实现这个 sleep() 函数让代码暂停指定 ...

  7. 洛谷——P2236 [HNOI2002]彩票

    P2236 [HNOI2002]彩票 给你$m$个数,从中挑$n$个数,使得这$n$个数的倒数之和恰好等于$\frac{x}{y}$ 常见的剪纸思路: 如果当前的倒数和加上最小可能的倒数和$>$ ...

  8. 类与类之间的关系UML模型图

    关联.依赖.聚合.组合.泛化.实现 类之间可能存在以下几种关系:关联(association).依赖(dependency).聚合(Aggregation,也有的称聚集).组合(Composition ...

  9. Yii 时间日期组件与composer 下载中出现的问题

    首先本篇主要讲3点 一个Yii时间日期组件的两种用法 笔者使用composer下载该组件时出现问题的解决办法 1.composer下载出现的问题 file could not be downloade ...

  10. linux设置系统时间与各种阻塞

    前阵子做了一个P2P的通信系统,发现开机的时候和中间运行的时候会莫名报错,这个问题找了好久,后来从日志中看出来,所有节点上阻塞的操作同时超时. 而在超时左右,有新节点自动加入系统. 在新节点加入系统的 ...