PAT Advanced—1147 Heaps (30分)

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：
  判断是否为大根堆，小根堆，或者不是堆

思路：

1. 直接，完全二叉树的性质，遍历一次，判断是否为堆，用每个结点和父节点比较，方便一些，即v[j/2] > v[j] 这样。
2. 层次遍历转后序遍历，直接按照后序遍历的访问方式递归，递归的值是层次遍历结点的位置2j为左子树，2j+1为右子树
3. 不输出末尾的空格可以采取printf("%d%s", v[j], x==1? “\n”: " "); 这样的方式判断，很方便

代码：(C++)

#include
#include

using namespace std;

int m, n;
vector<int> v;

void postTraversal(int x)
{
	if(x > n)
		return;
	postTraversal(x*2);
	postTraversal(x*2+1);
	printf("%d%s", v[x], x == 1? "\n": " ");
}

int main()
{
	
	cin>>m>>n;
	v.resize(n+1);
	for(int i=0; i<m; i++)
	{
		int min_s = 1, max_s = 1;
		for(int j=1; j<=n; j++)
			cin>>v[j];
		for(int j=2; j<=n; j++)
		{
			//用父节点来判断更方便
			if(v[j/2] > v[j]) min_s = 0;
			if(v[j/2] < v[j]) max_s = 0;
			
//			if( 2*j < n && (v[j] < v[2*j] || v[j] < v[2*j+1]))
//				max_s = 0;  //此时不再是大根堆
//			if( 2*j < n && (v[j] > v[2*j] || v[j] > v[2*j+1]))
//				min_s = 0; //此时不是小根堆
		}
		if(max_s == 0 && min_s == 0)
			printf("Not Heap\n");
		else if(max_s == 1)
			printf("Max Heap\n");
		else if(min_s == 1)
			printf("Min Heap\n");
		postTraversal(1);
	}
	
	return 0;
}