PAT 1064 Complete Binary Search Tree[二叉树][难]

1064 Complete Binary Search Tree (30)（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

 题目大意：给出一组数，是一颗完全二叉树并且是二叉搜索树，输出层次遍历序列。

 //看到题目感觉很懵，完全没有做过这样的题。

 代码来自：https://www.nowcoder.com/questionTerminal/ab6b7429f9c34880837cf4830b2434df

#include<cstdio>
#include<algorithm>
using namespace std;
int node[1010],tree[1010];
int N,pos;
void creatTree(int root){//就相当于顺序着把根安排在了tree上。
    //递归安排上了之后自然就是层次存储。
    if(root>N)    return;
    int lchild=root*2,rchild=root*2+1;
    creatTree(lchild);
    tree[root]=node[pos++];
    creatTree(rchild);
}
bool cmp(int a,int b){
    return a<b;
}
int main(){
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        scanf("%d",&node[i]);
    }
    sort(node,node+N,cmp);//既然是二叉搜索树，那么排序完了之后自然就是中序遍历。
    pos=0;
    creatTree(1);
    for(int i=1;i<=N;i++){//注意下标是从1开始计算的。
        printf("%d",tree[i]);
        if(i<N){
            printf(" ");
        }
    }
}

1.根据完全二叉树在数组中的存储特点，下标的特点。左子树是root*2,右子树是root*2+1.

2.使用pos来记录下标，因为是中序，所以是中序遍历存完全二叉树。

3.使用二叉搜索树的性质，中序遍历即从小到大。