1155 Heap Paths (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))

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

思路

• 层次遍历存在数组里即可，无需建树，当前结点编号d，则其左右节点编号为2d和2d+1（编号为1~N时）
• 输出使用深度优先搜索DFS，先遍历右子树再遍历左子树，不要忘记回溯

Problem

• “Not Heap”写成"No Heap"，也因此验证了一半的测试样例是"Not Heap"
• 忘记添加根节点路径
• 从根节点往下判断节点是否大于或小于其父节点，反向判断会稍微复杂

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1010;
int keys[maxn]; //层序遍历的节点
bool isMax = 1, isMin = 1;
vector<int> path;
int n;

//直接在数组中dfs
void dfs(int d)
{
    if(2 * d > n){ //不再往下搜索
        if(d <= n){
            for(int i = 0; i < path.size(); ++i){
                if(i != path.size() - 1)
                    printf("%d ",path[i]);
                else printf("%d\n",path[i]);
            }
        }
        return;
    }
    //先走右子树
    path.push_back(keys[2 * d + 1]);
    dfs(2 * d + 1);
    path.pop_back();//回溯
    //再走左子树
    path.push_back(keys[2 * d]);
    dfs(2 * d);
    path.pop_back();//回溯
}

int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i){
        scanf("%d",&keys[i]);
    }
    path.push_back(keys[1]);//别忘了路径中添加根节点
    dfs(1); //从鞥节点开始搜素
    for(int i = 2; i <= n; ++i){
       if(keys[i / 2] > keys[i]) isMin = 0;
       if(keys[i / 2] < keys[i]) isMax = 0;
    }
    if(isMax) printf("Max Heap");
    else if(isMin) printf("Min Heap");
    else printf("Not Heap");
    return 0;
}

标签：Paths,Heap,10,1155,30,tree,Sample,98,heap
来源： https://blog.csdn.net/weixin_36313227/article/details/90339686