PAT甲级1022

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.

翻译：一个二叉搜索树递归地表示为带有以下特点的二叉树

①左子树的节点的值都小于当前节点的值

②右子树的节点的值都大于等于当前节点的值

左子树和右子树也同时符合二叉搜索树的特点

一个完全二叉树是完全填充的树，除了最底层，是从左到右填充

现在给定一系列不同的非负整数值，如果需要树也必须是CBT，则可以构造唯一的BST。您应该输出这个BST的层序遍历序列。

过程：

①先构造一个二叉树，利用完全二叉树的性质

void creatTree(int root)
{
    if(root>N) return ;
    int lchild = root * 2, rchild = root *2 + 1;
    creatTree(lchild);
    tree[root] = node[pos++];   //利用数组存储树的节点
    creatTree(rchild);
}

将输入的测试的数据进行排序，从小到大，再依次中序放入数组，最后对数组进行输出，即可得到所要的输出。

完整代码段

#include <iostream>
#include<algorithm>
using namespace std;
int tree[1010];
int node[1010];
int pos = 1;
void creatTree(int root, int N)
{
    if (root > N) return ;
    int lchild = root * 2;
    int rchild = root * 2 + 1;
    creatTree(lchild,N);
    tree[root] = node[pos++];
    creatTree(rchild,N);
}

bool cmp(int a, int b){
    return a<b;
}

int main()
{
    int N;
    cin>>N;
    for(int i=1;i<=N;i++)
        cin>>node[i];
    sort(node+1,node+N+1,cmp);
    
    creatTree(1,N);
    cout<<tree[1];
    for(int j = 2;j <= N;j++)
        cout<<" " <<tree[j];
}