反转了链表的代码 哪里错了啊？

Question

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
         ListNode *temp=NULL;
         ListNode *p=pHead;
         ListNode *q=pHead->next;
         while(p!=NULL){
             p->next=temp;
             temp=p;
             p=q;
             q=q->next;
         }
        return p;
    }
};

Answer 1

一个节点时q=q->next会报错

达到链表末尾时
q=NULL;
q=q->next;报错Segmentation fault

while(p != NULL)
{
     p->next=temp;
     temp=p;
     p=q;
     if(q != NULL)
     {   
         q=q->next;
     }
 }
return temp;

循环的出口条件不正确，判断p=NULL是退出循环，return p;返回结果为NULL
到达末尾节点时p,q,q->next都为空，temp指向最后一个节点，即新的头节点

//迭代实现  
struct node * reverse_it(struct node *pHead)  
{  
    struct node *pNewHead = NULL;  
    struct node *pPrev = NULL;  
    struct node *pCur = pHead;  
    while(pCur != NULL)  
    {  
        struct node *pTmp = pCur->pNext;  
        if(pTmp == NULL)  
        {  
            pNewHead = pCur;  
        }  
        pCur->pNext = pPrev;  
        pPrev = pCur;  
        pCur = pTmp;  
    }  
  
    return pNewHead;  
}