1147 Heaps
1147 Heaps(30 分)
Input Specification:
Output Specification:
Sample Input:
Sample Output:
#include <cstdio>
int queryCnt,n;
];
void postOrderTraversal(int root)
{
    if(root>n) return;
    postOrderTraversal(root*);
    postOrderTraversal(root*+);
    printf("%d",CBT[root]);
    ) printf("\n");//因为是后序遍历,最后访问根节点,而根节点的下标固定为1
    else printf(" ");
}
int main()
{
    scanf("%d%d",&queryCnt,&n);
    ;i<queryCnt;i++){
        ;j<=n;j++)
            scanf("%d",&CBT[j]);
        //判断,层序遍历完全二叉树的非叶结点
        ]>CBT[]?:;//大顶堆标记为1,小顶堆标记为0。题目保证结点个数至少有两个。
        ;j<=n/;j++){
            ){
                ] || (j*+<=n && CBT[j]<CBT[j*+])){
                    flag=-;
                    break;
                }
            }
            ){
                ] || (j*+<=n && CBT[j]>CBT[j*+])){
                    flag=-;
                    break;
                }
            }
        }
        ) printf("Max Heap\n");
        ) printf("Min Heap\n");
        else printf("Not Heap\n");
        //输出后续序列
        postOrderTraversal();
    }
    ;
}
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