1064 Complete Binary Search Tree -PAT甲级

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本文介绍了一种特殊二叉搜索树——完全二叉搜索树（CBT）的构造方法，并通过给出的序列创建相应的CBT，最后输出该树的层次遍历序列。文章提供了详细的代码实现，帮助读者理解如何将中序排序转化为层次遍历的过程。

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题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

10
1 2 3 4 5 6 7 8 9 0

输出例子:

6 3 8 1 5 7 9 0 2 4

将一串数字按照从小到大的顺序排列就是中序排序，本题主要考察如何将中序排序转化为层次遍历，先访问根节点，再依次访问左右结点

此模板应熟记于心。

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int a[N],tree[N];
int n;
void inorder(int root,int &id){
	if(root>n) return;
	inorder(2*root,id);
	tree[root]=a[id];
	id++;
	inorder(2*root+1,id);
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	sort(a+1,a+n+1);
	int idd=1,root=1;
	inorder(root,idd);
	for(int i=1;i<=n;i++){
		if(i!=1) cout<<" ";
		cout<<tree[i];
	}
	return 0;
}