1064 Complete Binary Search Tree （30 分）

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<bits/stdc++.h>
using namespace std;
vector<int>inorder,level;
void levelorder(int start,int end,int index)
{
	if(start>end) return;
	int n=end-start+1;
	int l=log(n+1)/log(2);
	int leave=n-(pow(2,l)-1);
    int root = start + (pow(2, l - 1) - 1) + min((int)pow(2, l - 1), leave);//求左子樹的結點個數
	level[index]=inorder[root];
	levelorder(start,root-1,2*index+1);
	levelorder(root+1,end,2*index+2);
}
int main(void)
{
	int n;
	scanf("%d",&n);
	inorder.resize(n);
	level.resize(n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&inorder[i]);
	}
	sort(inorder.begin(),inorder.end());
	levelorder(0,n-1,0);
	printf("%d",level[0]);
	for(int j=1;j<n;j++)
	{
		printf(" %d",level[j]);
	}
	return 0;
}