1064. Complete Binary Search Tree (30)

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本文介绍了一种构建完全二叉搜索树的方法，并给出了相应的输入输出规范及示例。通过递归构建并进行层次遍历输出，确保了树的完整性和搜索特性。

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
using namespace std;

int N;
queue<int> q1;
void A(int i)
{
    if(i>N)
        return;
    A(i*2);
    q1.push(i);
    A(i*2+1);
}

int main()
{
    freopen("in.txt","r",stdin);

    cin>>N;
    int tN=N;
    vector<int> a;
    a.push_back(-999);
    while(tN--)
    {
        int t1;
        cin>>t1;
        a.push_back(t1);
    }

    sort(a.begin(),a.end());


    int tree[N+5];
    int k=1;
    A(1);
    while(!q1.empty())
    {
        int t2=q1.front();
        tree[t2]=a[k];
        k++;
        q1.pop();
    }

    int i;
    printf("%d",tree[1]);
    for(i=2;i<=N;i++)
    {
        printf(" %d",tree[i]);
    }

    return 0;
}