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
看了网上的做法，确实难想，利用二叉搜索树中序遍历是递增序列的性质，进行中序遍历的同时存进带有层序下标的数组中，最后将数组输出。
要我想的话肯定是建树后再层序输出，原谅我太愚蠢，懒得打了
https://blog.csdn.net/weixin_41144066/article/details/102591085 代码和和这个差不多~
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>s1, s2; int n;
void getlevel(int root)
{
s1.resize(n);
static int index = 0; //注意index要申明为全局变量
if (root >= n) return;
getlevel(root * 2 + 1);
s1[root] = s2[index++];
getlevel(root * 2 + 2);
}
int main()
{
cin >> n; s2.resize(n);
for (int i = 0; i < n; i++)
cin >> s2[i];
sort(s2.begin(), s2.end());
getlevel(0);
cout << s1[0];
for (int i = 1; i < n; i++)
cout << " "<<s1[i];
}