一、题目描述
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:
8
98 72 86 60 65 12 23 50
Sample Output 1:
98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap
Sample Input 2:
8
8 38 25 58 52 82 70 60
Sample Output 2:
8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap
Sample Input 3:
8
10 28 15 12 34 9 8 56
Sample Output 3:
10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap
二、解题思路
完全二叉树的dfs以及判断，我们依然可以使用两个全局变量isMax，isMin表示是最大堆，是最小堆，因为我们知道每次dfs到叶子结点的时候，形成的路径一定是单调的，所以直接比较最后两个元素即可。详情见代码注释。
三、AC代码
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int all[1010], N, isMax = 1, isMin = 1;
vector<int> path;
void dfs(int root) //dfs
{
if(root*2+1 >= N)
{
for(int i=0; i<path.size(); i++)
{
printf("%d%s", path[i], i != path.size()-1 ? " " : "\n");
if(i>0 && path[i] > path[i-1]) isMax = 0; //下一层的数字更大，则不是最大堆
if(i>0 && path[i] < path[i-1]) isMin = 0; //下一层的数组更小，则不是最小堆
}
}
else
{
if(root*2 + 2 < N) //如果还有右孩子，继续往下遍历
{
path.push_back(all[root * 2 + 2]);
dfs(root*2 + 2);
path.pop_back();
}
path.push_back(all[root*2 + 1]);
dfs(root*2 + 1);
path.pop_back(); //注意这里一定要pop_back
}
}
int main()
{
scanf("%d", &N);
for(int i=0; i<N; i++)
scanf("%d", &all[i]);
path.push_back(all[0]); //先将根结点push进path中
dfs(0);
if(isMin == 1) printf("Min Heap");
else isMax == 1 ? printf("Max Heap") : printf("Not Heap");
return 0;
}