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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…
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 1:898 72 86 60 65 12 23 50Sample Output 1:98 86 2398 86 1298 72 6598 72 60 50Max Heap
Sample Input 2:88 38 25 58 52 82 70 60Sample Output 2:8 25 708 25 828 38 528 38 58 60Min Heap
Sample Input 3:810 28 15 12 34 9 8 56Sample Output 3:10 15 810 15 910 28 3410 28 12 56Not Heap
题目重点信息提取：1. 输入：positive integer N 正整数，N distinct integer keys，N 个互不相等 的整数
给出一颗完全二叉树 **level order** traversal sequence of a **complete binary tree**
提取重点翻译：level order 与 complete binary tree，input 一个完全二叉树的层序遍历序列
2. 输出：重点翻译：first print **all the paths from the root to the leaves**
先打印出所有从根结点到叶子结点的路径，all the paths in its **right subtree** must be printed **before** those in its **left subtree**，到右子树的路径要先于到左子树路径打印。
画图对应样例的输入输出也可以快速判断出来
思路：深度遍历并打印出所有的路径（先右后左），用 vector 存储路径上的所有结点，通过 push 和 pop 回溯，维护路径，关于 index <= n，由于是先右后左，需要对只有左叶子结点而无右叶子结点的点进行特判。
#include <iostream>
#include <stdio.h>
#include <vector>
using namespace std;
int n,a[1001],isMaxHeap = 1,isMinHeap = 1;
vector<int> v;
void R_dfs(int index){// 从右至左的深度优先遍历
if(index * 2 > n && index * 2 + 1 > n){
if(index <= n){// 由于是先右后左，需要对只有左叶子结点而无右叶子结点的点进行特判
for(int i = 0; i < v.size(); i++)
printf(“%d%s”,v[i], i != v.size()-1 ? ” ” : “\n”);
}
}
else{
v.push_back(a[index *2 + 1]); // 深度遍历右子树
R_dfs(index * 2 + 1);
v.pop_back();
v.push_back(a[index * 2]); // 深度优先遍历左子树
R_dfs(index * 2);
v.pop_back();
}
}
int main()
{
scanf(“%d”,&n);
for(int i = 1; i <= n; i++){// 这里从 i = 1 开始，方便后续对二叉树有无右子树进行判断
scanf(“%d”,&a[i]);
}
v.push_back(a[1]);
R_dfs(1);
for(int i = 2; i <= n; i++){// 判断大小顶堆
if(a[i/2] > a[i])
isMinHeap = 0;
else if(a[i/2] < a[i])
isMaxHeap = 0;
}
if(isMinHeap == 1)
printf(“%s\n”,”Min Heap”);
else
printf(“%s\n”,isMaxHeap == 1 ? “Max Heap” : “Not Heap”);
return 0;
}
