04-树6 Complete Binary Search Tree（30 分）

题目来源：中国大学MOOC-陈越、何钦铭-数据结构-2018春
作者: 陈越
单位: 浙江大学
问题描述：
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.
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

解答：
存储数据的结构均选择数组，因为完全二叉树用数组更好做（可直接利用性质）。
首先，因为搜索二叉树左子节点<根<右子节点，发现若给n个数排好序，则这个序列是完全搜索二叉树的中序遍历。要建立这棵树，光知道中序遍历是不够的，因此引入完全二叉树的性质，就是给每个节点按1-N编号，则设根节点为i，左节点为2i,右子节点为2i+1，这样就可以从根节点1开始中序遍历完全二叉搜索树，在遍历的同时建树。中序遍历是访问节点时cout，那么现在需要建树，我们需要给这个节点赋一个值，这个值就是排好序的n个数依次取出即可。pos变量就是模拟现在取到哪个数了，因此要设置为全局。
注意递归建树的return条件是发现根的值大于N时返回。然后因为完全二叉树编号即是按层从上到下从左到右编号，所以输出直接按序输出Tree数组里的N个数即可。

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=1001;
int N;
int input[maxn],tree[maxn],pos=0;
void inputData()
{
    cin>>N;
    for(int i=0;i<N;i++)
    {
        cin>>input[i];
    }
}
void buildTree(int root)
{
    int leftChild=2*root,rightChild=2*root+1;
    if(root>N)
        return;
    buildTree(leftChild);
    tree[root]=input[pos++];
    buildTree(rightChild);
}
void output()
{
    for(int i=1;i<=N;i++)
    {
        if(i!=N)
            cout<<tree[i]<<" ";
        else
            cout<<tree[i];
    }
}
int main()
{
    inputData();
    sort(input,input+N);
    buildTree(1);
    output();
    return 0;
}