PAT_A1064#Complete Binary Search Tree

Source:

PAT A1064 Complete Binary Search Tree (30 分)

Description:

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 (≤). 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

Keys:

• 完全二叉树（Complete Binary Tree）
• 二叉查找树（Binary Search Tree）
• 二叉树的建立
• 二叉树的遍历

Attention:

• 很好的一道题，全面考察了完全二叉树和二叉查找树的性质

Code:

/*
Data: 2019-06-26 16:48:26
Problem: PAT_A1064#Complete Binary Search Tree
AC: 13:22

题目大意：
BST定义：lchild < root <= rchild
给定一个序列，建立CBT和BST，打印层次遍历

基本思路：
一维数组作为CBT的存储结构，结点序号从1~N;
中序遍历CBT，结果递增有序，因此遍历的同时依次赋值排序好的键值即可；
CBT树的遍历：左子树=i*2，右子树=i*2+1，空树i>n
CBT的层次遍历：打印cbt[1]~cbt[n]即可
*/
#include<cstdio>
#include<queue>
using namespace std;
const int M=1e3+10;
int cbt[M],n,x;
priority_queue<int,vector<int>,greater<int> > bst;

void InOrder(int root)
{
    if(root > n)
        return;
    InOrder(root*2);
    cbt[root] = bst.top();
    bst.pop();
    InOrder(root*2+1);
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE

    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        scanf("%d", &x);
        bst.push(x);
    }
    InOrder(1);
    for(int i=1; i<=n; i++)
        printf("%d%c", cbt[i],i==n?'\n':' ');

    return 0;
}