PAT 1064 Complete Binary Search Tree(30分)

1064Complete Binary Search Tree(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 integerN(≤1000). ThenNdistinct 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

思路

• 使用數組構造徹底二叉樹。
• 一棵徹底二叉樹能夠構造出惟一的一棵徹底二叉查找樹（CBST）。
• 二叉查找樹（BST）的中序序列是遞增序列，因而在徹底二叉樹的中序遍歷中構造出中序序列的CBST。最後將CBT順序輸出便是CBST的層次遍歷序列。

代碼

#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 1010;

int n, number[maxn], cbt[maxn];
int index = 0;

void inOrder(int root)
{
    if (root > n) return;
    inOrder(root * 2);
    cbt[root] = number[index ++];
    inOrder(root * 2 + 1);
}

int main()
{
    scanf("%d", &n);
    
    for (int i = 0; i < n; i ++)
    {
        scanf("%d", &number[i]);
    }
    sort(number, number + n);
    inOrder(1);
    for (int i = 1; i <= n; i ++)
    {
        printf("%d", cbt[i]);
        if (i < n)
            printf(" ");
    }
    return 0;
}