1147 Heaps (30 分)——甲级（堆）

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure))

Your job is to tell if a given complete binary tree is a heap.

Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.

Output Specification:
For each given tree, print in a line Max Heap if it is a max heap, or Min Heap for a min heap, or Not Heap if it is not a heap at all. Then in the next line print the tree’s postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.

Sample Input:
3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56
Sample Output:
Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10

题目大意：给出M个询问，每个询问给出N个数字，表示完全二叉树的层序遍历。判断这棵树是否是堆，最后输出后序遍历。
思路：对于每个有孩子的结点，去判断它是否比孩子结点大于等于或者小于等于：如果比孩子节点小，肯定不是最大堆；如果比孩子结点大，肯定不是最小堆。最后输出后序遍历。

代码如下：

#include <iostream>
#include <vector>
using namespace std;
vector<int>v;/* 存每次询问的后序遍历结果 */
void postorder(int a[], int index, int n)
{
    if(index > n) return ;
    postorder(a, index*2, n);
    postorder(a, index*2+1, n);
    v.push_back(a[index]);
}
int main()
{
    int m, n;
    cin >> m >> n;
    while(m--)
    {
        int i, a[1001];
        for(i=1; i<=n; i++) cin >> a[i];
        int is_min = 1, is_max = 1;/* 先假设是最大堆或者最小堆 */
        for(i=1; i*2<=n; i++)/* 如果有左孩子就进入循环 */
        {
             if(a[i] < a[i*2]) is_max = 0;
             if(a[i] > a[i*2]) is_min = 0;
             if(i*2+1 <= n)/* 如果有有孩子，则还需要判断右孩子 */
             {
                 if(a[i] < a[i*2+1]) is_max = 0;
                 if(a[i] > a[i*2+1]) is_min = 0;
             }
        }
        if(is_max == 1) cout << "Max Heap\n";
        else if(is_min == 1) cout << "Min Heap\n";
        else cout << "Not Heap\n";
        postorder(a, 1, n);
        cout << v[0];
        for(i=1; i<v.size(); i++)
            cout << " " << v[i];
        cout << endl;
        v.clear();
    }
    return 0;
}

总结

1. 一棵二叉树的层序遍历可以获得的信息：建树、前序、中序、后序。
2. 在求解问题的时候，如果要判断是否满足一个性质，而这个性质又需要局部满足性质，则可以先假定满足这个性质，然后通过访问局部去判断，只有有一处不满足就整体不满足这个性质了。
   如：在本题中，堆的局部也是堆，每个结点又必须既比左孩子大于等于（小于等于），又比右孩子大于等于（小于等于）。所以在判断是否是堆是，先假设是堆，然后通过遍历每个结点，只有有一处不满足，相应变量就置为0。最后去判断就可以了。