Complete Binary Search Tree

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：根据完全二叉树性质找规律

 解题感悟：1、记住vector中的sort和cmath文件中的一些函数很重要

               2、注意特殊情况下输出的格式（本题的特殊情况是只有一个输入的情况）

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath> 
using namespace std;
int main(){
    int n;
    cin >> n;
    vector<int> v;
    for (int i = 0; i < n; i++)
    {
        int number;
        cin >> number;
        v.push_back(number);
    }
    sort(v.begin(), v.end());
    int zhishu = log(n)/log(2);
    int tichushu = n -pow(2,zhishu)+ 1;
    vector<int> v1;
    for (int i = 0; i < tichushu;i++)
        v1.push_back(v[i * 2]);
    for (int i = 0; i < tichushu; i++)
        v.erase(v.begin() + i);
    int flag = 1;
    for (int i = 1; i <= zhishu;i++)
    {
        for (int j = 0; j < pow(2,i-1);j++){
            if (flag == 1)
            {
                cout << v[v.size() / pow(2, i) + j * v.size() / pow(2, i-1)];
                flag = 0;
            }
            else
            {
                cout << " " << v[v.size() / pow(2, i) + j * v.size() / pow(2, i-1)];
            }    
        }
    }
    for (int i = 0; i < v1.size();i++)
    {
        if (flag == 1)
        {
            cout<< v1[i];
            flag = 0;
        }
        else
        cout << " " << v1[i];
    }
    return 0;
}