题目粗心:
现有M个查问,每一个查问给定一个长度为N的齐全二叉树层序序列,判断该二叉树是大根堆,小根堆和非堆,而后输入该齐全二叉树的后序遍历序列。
算法思路:
对于齐全二叉树能够应用一个数组来保留其层序序列,而后应用函数isMaxHeap和isMinHeap别离判断该齐全二叉树是否是大根堆还是小根堆,如果都不是则输入Not Heap,而后再应用postTraverse对该齐全二叉树进行后序遍历,拜访节点的时候输入节点即可。
isMaxHeap
// 判断是否是大根堆
bool isMaxHeap(){
for (int i = 1; i <= N; ++i) {
if((2*i<=N&&heap[2*i]>heap[i])||(2*i+1<=N&&heap[2*i+1]>heap[i])){
return false;
}
}
return true;
}
isMinHeap
// 判断是否是小根堆
bool isMinHeap(){
for (int i = 1; i <= N; ++i) {
if((2*i<=N&&heap[2*i]<heap[i])||(2*i+1<=N&&heap[2*i+1]<heap[i])){
return false;
}
}
return true;
}
提交后果:
AC代码:
#include<cstdio>
using namespace std;
int heap[1005];
int M,N;
// 判断是否是大根堆
bool isMaxHeap(){
for (int i = 1; i <= N; ++i) {
if((2*i<=N&&heap[2*i]>heap[i])||(2*i+1<=N&&heap[2*i+1]>heap[i])){
return false;
}
}
return true;
}
// 判断是否是小根堆
bool isMinHeap(){
for (int i = 1; i <= N; ++i) {
if((2*i<=N&&heap[2*i]<heap[i])||(2*i+1<=N&&heap[2*i+1]<heap[i])){
return false;
}
}
return true;
}
int num;// 管制空格输入
void postTraverse(int root){
if(root>N) return;
postTraverse(2*root);
postTraverse(2*root+1);
printf("%d",heap[root]);
if(num<N-1) printf(" ");
++num;
}
int main(){
scanf("%d %d",&M,&N);
for (int i = 0; i < M; ++i) {
// M次查问
num = 0;// 每次都得赋值为0
for (int j = 1; j <= N; ++j) {
scanf("%d",&heap[j]);
}
if(isMinHeap()){
printf("Min Heap\n");
} else if(isMaxHeap()){
printf("Max Heap\n");
} else {
printf("Not Heap\n");
}
postTraverse(1);
printf("\n");// 记得换行
}
return 0;
}
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