1155 Heap Paths (30point(s)) - C语言 PAT 甲级

1155 Heap Paths (30point(s))

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))

One thing for sure is that all the keys along any path from the root to a leaf in a max/min heap must be in non-increasing/non-decreasing order.

Your job is to check every path in a given complete binary tree, in order to tell if it is a heap or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (1

Output Specification:

For each given tree, first print all the paths from the root to the leaves. Each path occupies a line, with all the numbers separated by a space, and no extra space at the beginning or the end of the line. The paths must be printed in the following order: for each node in the tree, all the paths in its right subtree must be printed before those in its left subtree.

Finally 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.

Sample Input:

8
98 72 86 60 65 12 23 50

Sample Output:

98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap

Sample Input:

8
8 38 25 58 52 82 70 60

Sample Output:

8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap

Sample Input:

8
10 28 15 12 34 9 8 56

Sample Output:

10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap

题目大意：

输入 N 个节点的完全二叉树，

输出从根节点到所有叶子点的路径，输出顺序由右到左，并判断该树是大顶堆或小顶堆或不是堆

设计思路：

• 按照由右到左 dfs 遍历完全二叉树，队列存储路径，遇到叶子节点输出路径
• 判断大小顶堆，父节点一直比子节点大为大顶堆，父节点一直比子节点小为小顶堆，否则不是堆

编译器：C (gcc)

#include 

int n, tree[1010];
int path[1010] = {
     0}, cnt = 0;

void dfs(int root)
{
     
        if (root > n)
                return ;
        path[cnt++] = tree[root];
        if (root * 2 > n) {
     
                int i;
                printf("%d", path[0]);
                for (i = 1; i < cnt; i++)
                        printf(" %d", path[i]);
                printf("\n");
                cnt--;
                return ;
        }
        dfs(root * 2 + 1);
        dfs(root * 2);
        cnt--;
}

void is_heap()
{
     
        int min = 1, max = 1;
        int i;
        for (i = 2; i <= n; i++) {
     
                if (tree[i / 2] > tree[i])
                        min = 0;
                else if (tree[i / 2] < tree[i])
                        max = 0;
        }
        if (min == 1)
                printf("Min Heap");
        else if (max == 1)
                printf("Max Heap");
        else
                printf("Not Heap");
        return ;
}

int main(void)
{
     
        int i;

        scanf("%d", &n);
        for (i = 1; i <= n; i++)
                scanf("%d", &tree[i]);
        dfs(1);
        is_heap();
        return 0;
}