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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