我正在尝试使用C中的递归来反转链表，我对我的递归函数有一些疑问

Question

下面是程序

#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node *next;
};
struct node *head;
struct node* reverse_ll(struct node* hnode)
{
if(hnode == 0)
{
return 0;
}
if(hnode->next == 0)
{
head=hnode;
return hnode;
}
struct node* ptr=reverse_ll(hnode->next);
ptr->next=hnode;
hnode->next=0;
//return hnode;
}
void display()
{
struct node *ptr;
ptr=head;
if(ptr==0)
{
printf("empty");
}
else
{
while(ptr!=0)
{
printf("%d->",ptr->data);
ptr=ptr->next;
}
printf("null");
}
}
int main()
{
struct node* h;    
lastinsert(1);
lastinsert(2);
lastinsert(3);
lastinsert(4);
lastinsert(5);
display();
h=reverse_ll(head);
display();
return 0;
}

在函数reverse_ ll((中；返回hnode"；我得到了正确的输出。当我评论"ptr从哪里接收地址？"；返回hnode"；？

输出：1->2->3->4->5->无效的5->4->3->2->1->空

Answer 1

reverse_ll()在递归情况下必须返回struct node *：

#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node *next;
};
struct node *head;
void lastinsert(int data) {
struct node **c = &head;
for(; *c; c = &(*c)->next);
*c = malloc(sizeof(*head));
if (!(*c)) {
printf("malloc failedn");
return;
}
(*c)->data = data;
(*c)->next = NULL;
}
struct node *reverse_ll(struct node* hnode) {
if(!hnode)
return NULL;
if(!hnode->next) {
head = hnode;
return hnode;
}
struct node *ptr=reverse_ll(hnode->next);
ptr->next=hnode;
hnode->next = NULL;
return hnode;
}
void display() {
if(!head) {
printf("empty");
return;
}
for(struct node *ptr = head; ptr; ptr = ptr->next) {
printf("%d->",ptr->data);
}
printf("nulln");
}
int main() {
for(int i = 1; i <= 5; i++) {
lastinsert(i);
}
display();
reverse_ll(head);
display();
// It's good practice to implement a function that frees you list
// which you would call here.
return 0;
}

和示例运行：

$ ./a.out
1->2->3->4->5->null
5->4->3->2->1->null