1155.Heap Paths-PAT甲级真题（DFS+堆和二叉树的概念）

标签：heap   void   zed   其他   amp   搜索   extra   index   镜像   

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

-----------------------------------------------------------

注意：

完全二叉树的定义：若设二叉树的深度为h，除第h层外，其他各层（1~h-1）的节点数都达到最大个数，第h层所有节点都连续集中在最左边（就是说可能会有只有左子树没有右子树的情况，按照宽度优先遍历的时候搜索到空节点时树一定已经遍历完）。完全二叉树是由满二叉树而引出来的。对于深度为k的，由n个节点的二叉树，当且仅当其每一个节点的深度都与深度为k的二叉树中编号从1到n的节点一一对应称之为完全二叉树。

 题目大意：

　　给出一颗完全二叉树，打印从根节点开始到所有叶节点的路径，打印的顺序先右后坐，及先序遍历的镜像。然后判断是否是堆。

分析：

1.深度打印所有的路径（从右向左，及先序的镜像），vector保存一条路径上的所有节点，通过push和pop实现回溯，维护路径，index<=n是对只有左节点没有右节点的情况的特判。

2.判断是否为堆：从第二个节点开始遍历，如果比父节点小，就不是小顶堆，如果比父节点大，就不是大顶堆。

#include<stdio.h>
#include<vector>
#pragma warning(disable:4996)
#define maxn 1004
using namespace std;
int n;
int he[maxn];
vector<int>road;
void dfs(int index)
{
    if (index * 2 > n) {
        if (index <= n) {
            for (int i = 0; i < road.size(); i++) {
                printf("%d%s", he[road[i]], i == road.size()-1 ? "\n" : " ");
            }
        }//对只有左子树没有右子树的特判

    }
    else {
        road.push_back(index*2+1);
        dfs(index * 2 + 1);
        road.pop_back();
        road.push_back(index * 2);
        dfs(index * 2);
        road.pop_back();
    }
}
int main()
{
    scanf("%d",&n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &he[i]);
    }
    road.push_back(1);
    dfs(1);
    int ismax = 1, ismin = 1;
    for (int i = 2; i <= n; i++) {
        if (he[i / 2] > he[i])ismin = 0;
        if (he[i / 2] < he[i])ismax = 0;
    }
    printf("%s",ismin ==1?"Min Heap":(ismax == 1?"Max Heap":"Not Heap"));
    getchar();
    getchar();
    return 0;
}

1155.Heap Paths-PAT甲级真题（DFS+堆和二叉树的概念）

标签：heap   void   zed   其他   amp   搜索   extra   index   镜像   

原文地址：https://www.cnblogs.com/zxzmnh/p/11629929.html