PAT (Advanced Level) Practise 1064 Complete Binary Search Tree (30)

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本文介绍如何通过给定的一组非负整数构造一颗既为二叉搜索树又为完全二叉树的数据结构，并输出该树的层次遍历序列。

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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

CHEN, Yue

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

题意：给你n个数，让你将他们插入一棵完全二叉搜索树中

解题思路：很明显将所有数从小到大排序后，这个序列就是这棵树的中序遍历，dfs插入即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <bitset>
#include <set>
#include <vector>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,a[1005],cnt,b[1005];

void dfs(int k)
{
    if(k>n) return ;
    dfs(k*2);
    b[k]=a[cnt++];
    dfs(k*2+1);
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        cnt=1;
        dfs(1);
        printf("%d",b[1]);
        for(int i=2;i<=n;i++) printf(" %d",b[i]);
        printf("\n");
    }
    return 0;
}