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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