PAT 1155 Heap Paths （30 分）- 甲级

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

题目大意：给出一颗完全二叉树，打印出从根节点到所有叶节点的路径，打印顺序先右后左，即先序遍历的镜像。然后判断该树是大顶堆、小顶堆或者不是堆～

分析：1.深搜打印出所有路径（从右往左，即先序的镜像），vector保存一路上的节点，通过push和pop回溯，维护路径，index <= n是对只有左叶节点没有右叶节点的点特判
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 <iostream>
#include <vector>
using namespace std;
vector<int> v;
int a[1009], n, isMin = 1, isMax = 1;
void dfs(int index) {
    if (index * 2 > n && index * 2 + 1 > n) {
        if (index <= n) {
            for (int i = 0; i < v.size(); i++)
                printf("%d%s", v[i], i != v.size() - 1 ? " " : "\n");
        }
    } else {
        v.push_back(a[index * 2 + 1]);
        dfs(index * 2 + 1);
        v.pop_back();
        v.push_back(a[index * 2]);
        dfs(index * 2);
        v.pop_back();
    }
}
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    v.push_back(a[1]);
    dfs(1);
    for (int i = 2; i <= n; i++) {
        if (a[i/2] > a[i]) isMin = 0;
        if (a[i/2] < a[i]) isMax = 0;
    }
    if (isMin == 1)
        printf("Min Heap");
    else 
        printf("%s", isMax == 1 ? "Max Heap" : "Not Heap"); 
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

vector<int> v;

int a[1009], n, isMin = 1, isMax = 1;

void dfs(int index) {

if (index * 2 > n && index * 2 + 1 > n) {

if (index <= n) {

for (int i = 0; i < v.size(); i++)

printf("%d%s", v[i], i != v.size() - 1 ? " " : "\n");

}

} else {

v.push_back(a[index * 2 + 1]);

dfs(index * 2 + 1);

v.pop_back();

v.push_back(a[index * 2]);

dfs(index * 2);

v.pop_back();

}

int main() {

cin >> n;

for (int i = 1; i <= n; i++)

scanf("%d", &a[i]);

v.push_back(a[1]);

dfs(1);

for (int i = 2; i <= n; i++) {

if (a[i/2] > a[i]) isMin = 0;

if (a[i/2] < a[i]) isMax = 0;

}

if (isMin == 1)

printf("Min Heap");

else

printf("%s", isMax == 1 ? "Max Heap" : "Not Heap");

return 0;

}

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