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:
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题意：给出完全二叉树的层序遍历，要求出所有从根节点到叶子节点的路径（右子树优先），还要判断是否为堆，如果是判断是大顶堆还是小顶堆。
解题思路：完全二叉树的层序遍历，我们知道下标的关系，节点的左右子节点的关系是2*i+1、2*i+2（从0开始），求出根节点到叶子节点所有路径，深度优先搜索dfs，用vector来保存经过路径的节点，通过push_back()和push_pop()来控制路的节点删除插入。当2*index+1和2*index+2都大于等于n，就说明这个节点已经是叶子节点，现在要做的就是将这条路径的节点打印出来，这里要注意的是要考虑只有左结点的情况，所以在打印的前面加个条件（index<n)，若不是到叶子节点，那我们先将右结点压入vector中，然后对右结点递归，递归完接下来就要push_pop()，删除上一个叶子节点，继续将左子节点压入到vector中，然后对左子节点递归，递归完也删除上一个叶子节点。判断堆这方面，我们先假设都是大顶堆小顶堆，然后我们for循环，从第二个循环，如果v[i]>v[(i-1)/2]就将小顶堆置为false，相反将大顶堆置为false，然后其中一个为true，就说明是什么堆，如果都是false，说明不是堆，直接输出Not Heap。
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
int n;
vector<int>v,path;
bool Minheap=true,Maxheap=true;
void dfs(int index)
{
if(2*index+1>=n&&2*index+2>=n)
{
if(index<n){//考虑到只有一个左结点（要是不考虑，会给不存在的右结点随机数）
for(int i=0;i<path.size();i++)
printf("%d%s",path[i],i==path.size()-1?"\n":" ");
}
}
else{
path.push_back(v[2*index+2]);
dfs(2*index+2);
path.pop_back();
path.push_back(v[2*index+1]);
dfs(2*index+1);
path.pop_back();
}
}
int main(void)
{
scanf("%d",&n);
v.resize(n);
for(int i=0;i<n;i++)
{
scanf("%d",&v[i]);
}
path.push_back(v[0]);
dfs(0);
for(int i=1;i<n;i++)
{
if(v[i]<v[(i-1)/2]) Minheap=false;
if(v[i]>v[(i-1)/2]) Maxheap=false;
}
if(Minheap==true)
printf("Min Heap");
else
printf("%s",Maxheap==true?"Max Heap":"Not Heap");
return 0;
}