1064. Complete Binary Search Tree (30)

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：搜索二叉树的中序遍历是一个递增的序列。所以需要把输入的序列sort一下。还有一个完全二叉树的性质是：下标为i的节点，它的左儿子下标2*i，右儿子2*i+1。

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
#define max 1005
int root[max];
int val[max];
int n,m;
bool cmp(int a, int b){
	return a<b;
}
void build(int index){
	if(index>n)return;
	int left=index*2;
	int right=index*2+1;
	build(left);
	root[index]=val[m++];
	build(right);
}
int main(){
	scanf("%d",&n);
	int i,j;
	for(i=0;i<n;i++){
		scanf("%d",&val[i]);
	}
	sort(val,val+n,cmp);
	m=0;
	build(1);
	printf("%d",root[1]);
	for(i=2;i<n+1;i++){
		printf(" %d",root[i]);
	}
	return 0;
}