题目

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<N≤1,000), the number of keys in the tree. Then the next line 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, 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 1:

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2
8
98 72 86 60 65 12 23 50

Sample Output 1:

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2
3
4
5
98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap

Sample Input 2:

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2
8
8 38 25 58 52 82 70 60

Sample Output 2:

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2
3
4
5
8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap

Sample Input 3:

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2
8
10 28 15 12 34 9 8 56

Sample Output 3:

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2
3
4
5
10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap

题解

思路

  • 根本是考你堆的存储
  • 堆的存储是一个数组，第一个0下标为空。真正的根在下标1。
  • 这样就可以保证，任何下标乘以2就是它的左孩子，乘以2+1就是它的右孩子
  • 然后题目考你DFS和maxheap还是minheap的判断，分开来都不难
  • DFS遍历树就是维护一个路径，能右就右，然后能左就左，注意回溯。
  • 判断heap先判断前两项的关系，然后看后面每项和它的父亲比大小即可。

数据结构

  • nodes 是一个堆
  • path 记录DFS的路径
  • isMax 记录是不是大顶堆

算法

  • 照着思路走一遍就OK。

代码

  • 由于使用Python可以AC，因此只放了Python的题解。
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def dfs(node):
    # 已经是叶子节点了
    if 2 * node >= num_nodes:
        print(" ".join(map(str, path)))
        return
    # 有右节点的，先压右节点
    if 2 * node + 1 < num_nodes:
        path.append(nodes[2 * node + 1])
        dfs(2 * node + 1)
        path.pop()
    # 再压左节点
    if 2 * node < num_nodes:
        path.append(nodes[2 * node])
        dfs(2 * node)
        path.pop()


num_nodes = int(input()) + 1
nodes = [0] + list(map(int, input().split()))
path = [nodes[1]]
dfs(1)
isMax = nodes[2] < nodes[1]
if isMax:
    for i in range(3,num_nodes):
        if nodes[i] > nodes[i//2]:
            print("Not Heap")
            break
    else:
        print("Max Heap")
else:
    for i in range(3,num_nodes):
        if nodes[i] < nodes[i//2]:
            print("Not Heap")
            break
    else:
        print("Min Heap")