C: How to free nodes in the linked list?

Question

How will I free the nodes allocated in another function?

struct node {
    int data;
    struct node* next;
};

struct node* buildList()
{
    struct node* head = NULL;
    struct node* second = NULL;
    struct node* third = NULL;

    head = malloc(sizeof(struct node));
    second = malloc(sizeof(struct node));
    third = malloc(sizeof(struct node));

    head->data = 1;
    head->next = second;

    second->data = 2;
    second->next = third;

    third->data = 3;
    third->next = NULL;

    return head;
}  

I call the buildList function in the main()

int main()
{
    struct node* h = buildList();
    printf("The second element is %d\n", h->next->data);
    return 0;
}  

I want to free head, second and third variables.
Thanks.

Update:

int main()
{
    struct node* h = buildList();
    printf("The element is %d\n", h->next->data);  //prints 2
    //free(h->next->next);
    //free(h->next);
    free(h);

   // struct node* h1 = buildList();
    printf("The element is %d\n", h->next->data);  //print 2 ?? why?
    return 0;
}

Both prints 2. Shouldn't calling free(h) remove h. If so why is that h->next->data available, if h is free. Ofcourse the 'second' node is not freed. But since head is removed, it should be able to reference the next element. What's the mistake here?

Answer 1

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