c 语言怎么释放链表节点,C：如何释放链表中的节点？

我如何释放在另一个函数中分配的节点？

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;

}

我在main()中调用buildList函数

int main()

{

struct node* h = buildList();

printf("The second element is %d\n", h->next->data);

return 0;

}

我想释放头,第二和第三个变量.

谢谢.

更新：

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;

}

两个打印2.不应该调用(h)删除h.如果是这样,为什么h-> next->数据可用,如果h是空的.当然,“第二”节点没有被释放.但是由于头被删除,它应该能够引用下一个元素.这里有什么错误？