void fun1(struct node* head)
{
if(head == NULL)
return;

fun1(head-next);
printf("%d ", head-data);
}

Question

void fun1(struct node* head)
{
  if(head == NULL)
    return;
  
  fun1(head->next);
  printf("%d  ", head->data);
}

Anónimo · Accepted Answer

fun1() prints the given Linked List in reverse manner. For Linked List 1->2->3->4->5, fun1() prints 5->4->3->2->1.

Anónimo · Answer

This method will give Null pointer exception. 
When head=Null then only method returns and execution pointer will come on the line printf("%d ", head->data); 
But, at this point head=Null so, code is trying to print null->data which will throw exception.
If the 'if' condition is if(head->next == NULL) return; then linked list will be printed in reverse manner.

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