1155 Heap Paths (30分) 一道題複習dfs打印路徑+堆排序調整過程

 

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

 

題目大意：給出一個層序遍歷的完全二叉樹，要你打印所有到葉子節點的路徑（先右後左）+判斷這棵樹是大頂堆還是小頂堆還是根本就不是堆。 

首先dfs打印路徑：可以用vector存儲路徑，也可以用一維數組path[]存儲。

1.vector存儲路徑（不用維護表示層數的變量level，但是要在dfs返回的時候彈出一個元素，也就是父結點）

void dfs(int root){
    if(root>n)return;
    path.push_back(heap[root]);
    //如果是葉子節點就打印路徑
    if(root*2>n&&root*2+1>n){
        bool flag = false;
        for(int i = 0;i<=level;i++){
            if(flag==false){
                cout<<path[i];
                flag = true;
            }
            else{
                cout<<" "<<path[i];
            }
        }
        cout<<endl;
    }
    //先右後左
    dfs(root*2+1,level+1);
    dfs(root*2,level+1);
    //彈出父結點
    path.pop_back();
}

2.path數組存儲路徑

void dfs(int root,int level){
    if(root>n)return;
    path[level] = heap[root];
    if(root*2>n&&root*2+1>n){
        bool flag = false;
        for(int i = 0;i<=level;i++){
            if(flag==false){
                cout<<path[i];
                flag = true;
            }
            else{
                cout<<" "<<path[i];
            }
        }
        cout<<endl;
    }
    dfs(root*2+1,level+1);
    dfs(root*2,level+1);
}

然後是堆排序時調整的函數downAdjust：其實這題可以直接用i/2表示父結點，與孩子結點i對比，就可以看出是什麼堆。但是爲了複習堆調整，還是把調整函數寫一遍，體會一下堆是怎麼調整的。

本題採用的思路是從n/2號結點開始向樹的上層遍歷（因爲從n/2號結點開始纔有孩子結點），每次調用downAdjusth函數，檢查該結點是否符合大頂堆or小頂堆的要求

//檢查是否是大頂堆
bool downAdjustMax(int low,int high){
    //i一開始爲該結點編號，j爲其左孩子編號
    //high爲總結點數
    int i = low;
    int j = i*2;
    while(j<=high){
        //如果右孩子比較大，就把右孩子編號賦值給j
        if(j+1<=high&&heap[j]<heap[j+1]){
            ++j;
        }
        //如果不符合大頂堆要求，就return
        if(heap[i]<heap[j])return false;
        //否則接着向下檢查
        else {
            i = j;
            j = i*2;
        }
    }
    return true;
}

完整代碼：

#include <iostream>
#include <vector>
using namespace std;
int heap[1001],n;
vector<int> path;
//打印路徑
void dfs(int root,int level){
    if(root>n)return;
    path.push_back(heap[root]);
    if(root*2>n&&root*2+1>n){
        bool flag = false;
        for(int i = 0;i<=level;i++){
            if(flag==false){
                cout<<path[i];
                flag = true;
            }
            else{
                cout<<" "<<path[i];
            }
        }
        cout<<endl;
    }
    dfs(root*2+1,level+1);
    dfs(root*2,level+1);
    path.pop_back();
}
//檢查是否是大頂堆
bool downAdjustMax(int low,int high){
    int i = low;
    int j = i*2;
    while(j<=high){
        if(j+1<=high&&heap[j]<heap[j+1]){
            ++j;
        }
        if(heap[i]<heap[j])return false;
        else {
            i = j;
            j = i*2;
        }
    }
    return true;
}
//檢查是否是小頂堆
bool downAdjustMin(int low,int high){
    int i = low;
    int j = i*2;
    while(j<=high){
        if(j+1<=high&&heap[j]>heap[j+1]){
            ++j;
        }
        if(heap[i]>heap[j])return false;
        else {
            i = j;
            j = i*2;
        }
    }
    return true;
}
int main(){
    cin>>n;
    for(int i = 1;i<=n;i++){
        cin>>heap[i];
    }
    dfs(1,0);
    bool isMax= true,isMin = true;
    //從n/2號結點到根節點遍歷檢查
    for(int i = n/2;i>=1;i--){
        if(downAdjustMax(i,n)==false)isMax = false;
    }
    if(isMax==false){
        for(int i = n/2;i>=1;i--){
        if(downAdjustMin(i,n)==false)isMin = false;
        }
    }
    if(isMax)cout<<"Max Heap";
    else if(isMin)cout<<"Min Heap";
    else cout<<"Not Heap";
    return 0;
}