1147 Heaps (30分)
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
思考：
-
根據層序遍歷構造二叉樹的過程，我感覺使用數組要簡單點，而且題目要求的M也不大，開個大數組，下標從1表示到M，這樣結點孩子 父親的關係也就知道了，結點1的左孩子就是12 右孩子就是12+1
-
使用一個包含結點關係的數組 進行遞歸構造二叉樹，
void build(node *root,int pos) { if(pos*2<=M) { node *p=new node(); p->data=sn[pos*2]; root->lchild=p; build(p,pos*2); } if(pos*2+1<=M) { node *p=new node(); p->data=sn[pos*2+1]; root->rchild=p; build(p,pos*2+1); } } -
構造玩二叉樹之後就是判斷是否是大小堆，我也是用了遞歸，但是這個遞歸最終不是遇到Null返回了，遇到Null返回就太遲了，我們只是判斷結點和他左右結點的關係，所以返回條件設置成結點的左右孩子爲空。
-
然後再寫判斷大小堆的遞歸函數後面詳細處理部分，（此時進行到這一步的root 指向的結點 左右孩子不會全爲Null） 分開處理就好了，不過也需要注意細節
bool judgeMax(node *root) { if(root->lchild==NULL&&root->rchild==NULL) return 1; else if(root->rchild==NULL) { if(root->data>=root->lchild->data) { return judgeMax(root->lchild); } else return 0; } else { if(root->data>=root->lchild->data&&root->data>=root->rchild->data) { return judgeMax(root->lchild)&&judgeMax(root->rchild); } else return 0; } }
AC代碼
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <unordered_map>
#include <list>
using namespace std;
struct node
{
int data;
node *lchild,*rchild;
};
int sn[9999];
vector<int> vec;
int N,M;
void build(node *root,int pos)
{
if(pos*2<=M)
{
node *p=new node();
p->data=sn[pos*2];
root->lchild=p;
build(p,pos*2);
}
if(pos*2+1<=M)
{
node *p=new node();
p->data=sn[pos*2+1];
root->rchild=p;
build(p,pos*2+1);
}
}
void df(node *root)
{
if(root==NULL)
return;
df(root->lchild);
df(root->rchild);
vec.push_back(root->data);
}
bool judgeMax(node *root)
{
if(root->lchild==NULL&&root->rchild==NULL)
return 1;
else if(root->rchild==NULL)
{
if(root->data>=root->lchild->data)
{
return judgeMax(root->lchild);
}
else
return 0;
}
else
{
if(root->data>=root->lchild->data&&root->data>=root->rchild->data)
{
return judgeMax(root->lchild)&&judgeMax(root->rchild);
}
else
return 0;
}
}
bool judgeMin(node *root)
{
if(root->lchild==NULL&&root->rchild==NULL)
return 1;
else if(root->rchild==NULL)
{
if(root->data<=root->lchild->data)
{
return judgeMax(root->lchild);
}
else
return 0;
}
else
{
if(root->data<=root->lchild->data&&root->data<=root->rchild->data)
{
return judgeMin(root->lchild)&&judgeMin(root->rchild);
}
else
return 0;
}
}
int main()
{
cin>>N>>M;
for(int q=0; q<N; q++)
{
fill(sn,sn+9999,0);
for(int i=1; i<=M; i++)
{
cin>>sn[i];
}
node *root=new node();
root->data=sn[1];
build(root,1);
df(root);
bool maxx=judgeMax(root);
bool minn=judgeMin(root);
if(maxx)
cout<<"Max Heap"<<endl;
else if(minn)
cout<<"Min Heap"<<endl;
else
cout<<"Not Heap"<<endl;
for(int i=0; i<vec.size(); i++)
{
if(i==0)
cout<<vec[i];
else
cout<<' '<<vec[i];
}
cout<<endl;
vec.clear();
}
return 0;
}