C: How to free nodes in the linked list?

Simply by iterating over the list:

struct node *n = head;
while(n){
   struct node *n1 = n;
   n = n->next;
   free(n1);
}

An iterative function to free your list:

void freeList(struct node* head)
{
   struct node* tmp;

   while (head != NULL)
    {
       tmp = head;
       head = head->next;
       free(tmp);
    }

}

What the function is doing is the follow:

1. check if head is NULL, if yes the list is empty and we just return

2. Save the head in a tmp variable, and make head point to the next node on your list (this is done in head = head->next

3. Now we can safely free(tmp) variable, and head just points to the rest of the list, go back to step 1