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))
Your job is to tell if a given complete binary tree is a heap.
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.
For each given tree, 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. Then in the next line print the tree’s postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.
题意：给出完全二叉树的层次遍历，判断这棵完全二叉树是最大堆，最小堆，还是不是堆，再写出他的后序遍历。 思路：先讲层次遍历的结果都记录在vector中，然后首先比较v[0]和v[1]的结果如果是大于的话记flag=1，假设为大土堆，小于的话记flag=-1,假设为小土堆。然后从0遍历到(n - 1)/2（能取到）,如果在flag=1的情况下（说明假设是大土堆），如果出现v[j] < v[j * 2 + 1]和v[j2+2]并且j * 2 + 1 < n的时候则flag=0，说明不是堆。在flag=-1的情况下（说明假设是小土堆），如果出现 v[j * 2 + 1]和v[j2+2]并且j * 2 + 2 < n的时候则flag = 0,.说明不是堆。后序遍历注意的就是index>=n的临界条件的控制即可
#include<iostream> #include<cstdio> #include<stack> #include<queue> #include<vector> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<set> #include<map> #include<cctype> #include<cstdlib> #include<ctime> #include<unordered_map> using namespace std; int m,n; vector<int> v; void postorder(int index) { if(index >= n) return; postorder(index * 2 + 1); postorder(index * 2 + 2); printf("%d%s",v[index],index == 0 ? "\n" : " "); } int main() { scanf("%d%d",&m,&n); v.resize(n); for(int i = 0;i < m;i++) { for(int j = 0;j < n;j++) { scanf("%d",&v[j]); } int flag = v[0] > v[1] ? 1 : -1; for(int j = 0;j <= (n - 1) / 2;j++) { int left = j * 2 + 1,right = j * 2 + 2; if(flag == 1 && (v[j] < v[left] || (right < n && v[j] < v[right]))) flag = 0; if(flag == -1 && (v[j] > v[left] || (right < n && v[j] > v[right]))) flag = 0; } if(flag == 0) { printf("Not Heap\n"); } else { printf("%s Heap\n",flag == 1 ? "Max" : "Min"); } postorder(0); } return 0; }