PAT 甲级 1147 Heaps (30 分) (层序遍历，如何建树，后序输出，还有更简单的方法~)
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.
Input Specification:
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.
Output Specification:
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.
Sample Input:
3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56
Sample Output:
Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10
题意：
给一个树的层序遍历，判断它是不是堆，是大顶堆还是小顶堆。输出这个树的后序遍历～
题解：
在输入程序遍历的时候就借用队列建树，同时判断堆的类型，建好树以后递归进行后序遍历输出。
方法暴力过于繁琐，下面有别人家的代码。。。
本题思路虽然容易想，但是建树不太熟练，建树通常有两种方法：数组建树 和 链表建树，参见：建树的两种方法
AC代码：
#include<iostream> #include<algorithm> #include<queue> #include<stack> using namespace std; int m,n; struct node{ int v; node *l,*r; }; queue<node*>q; int sum=0; node* newnode(int x){//新建一个节点 node* newnode = new node; newnode->v=x; newnode->l=newnode->r=NULL; return newnode; } void postorder(node *&root){//后序遍历 if(root->l) postorder(root->l); if(root->r) postorder(root->r); sum++;//用于计算是否要加空格 cout<<root->v; if(sum!=n) cout<<" "; else cout<<endl; } int main(){ cin>>m>>n; int x; node *root; for(int i=1;i<=m;i++){ int f=0;//从小到大为1 while(!q.empty()) q.pop(); for(int j=1;j<=n;j++){ cin>>x; if(j==1){ root= newnode(x); q.push(root); continue; } while(!q.empty()){//层序遍历用队列更方便 node* a = q.front(); node* b = newnode(x); if(a->l==NULL){ a->l=b; }else if(a->r==NULL){ a->r=b; }else{ q.pop();//如果队头的节点的左右节点装满了就把它扔了再从队头拿出 continue; } q.push(b); if(f==0 && a->v<x){//判读树的类型 f=1;//从小到大为1 }else if(f==0 && a->v>x){ f=2;//从大到小为2 }else if(f==1 && a->v>x){ f=-1;//不符合条件为-1 }else if(f==2 && a->v<x){ f=-1; } break;//记得要跳出 } } sum=0; if(f==1){ cout<<"Min Heap"<<endl; }else if(f==2){ cout<<"Max Heap"<<endl; }else cout<<"Not Heap"<<endl; postorder(root); } return 0; }
更简单的不用建树的方法:
首先根据v[0]和v[1]的大小比较判断可能是大顶还是小顶，分别赋值flag为1和-1，先根据层序遍历，从0到n/2-1【所有有孩子的结点】判断他们的孩子是不是满足flag的要求，如果有一个结点不满足，那就将flag=0表示这不是一个堆。根据flag输出是否是堆，大顶堆还是小顶堆，然后后序遍历，根据index分别遍历index*2+1和index*2+2，即他们的左右孩子，遍历完左右子树后输出根结点，即完成了后序遍历～
#include <iostream> #include <vector> 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 / 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; }
