检查列表是否包含函数frontbacksplit中需要0或1个节点的测试？

Question

代码在此链接中 http://www.geeksforgeeks.org/merge-sort-for-linked-list/ 我的问题是，是否绝对需要在frontbacksplit（）中检查0或1个节点的代码片段，因为在调用它的mergesort（）函数中已经完成了对上述条件的检查...

我也试过自己上面代码的实现，但是我收到了一个错误，请你指出错误的位置以及你如何找到错误......

      /* Link list node */
struct node
{
    int data;
    struct node* next;
};


void mergesort(struct node** head);
void merge(struct node **head, struct node **a, struct node **b);
void split(struct node **head, struct node **a,struct node **b);

void split(struct node **head, struct node **a,struct node **b)
{
    /*i dont have to check if the list is empty or whether it has only one node in this function
    because that condition has already been checked in the calling function*/
    /*so i know for a fact that the list pointed to by head contains atleast 2 nodes*/
    struct node *front=*head;
    struct node *rear=(*head)->next;

    while(rear!=NULL)
    {
        rear=rear->next;
        if(rear!=NULL)
        {
            front=front->next;
            rear=rear->next;
        }
    }
    /*now front points to the end of the first list*/
    *a=*head;
    *b = front->next;
    front->next = NULL;
}


void mergesort(struct node** head)
{
    struct node *a = NULL;
    struct node *b = NULL;

    /*if list to be sorted is empty or contains only 1 element , then it is trivially sorted*/
    if((*head==NULL)||((*head)->next == NULL))
    {
        return;
    }

    /*partition the list into 2 halves*/
    split(head,&a,&b);


    mergesort(&a);
    mergesort(&b);

    merge(head,&a,&b);
}

void merge(struct node **head, struct node **a, struct node **b)
{
    struct node *x=*a;
    struct node *y=*b;
    struct node **s=NULL;
    struct node *z=NULL;

    /* x and y now point to the beginning of the 2 lists*/

    if((x==NULL)&&(y==NULL))    /*if both lists are empty*/
    {
        *head = NULL;
        return;
    }
    if((x==NULL))   /*if x is empty but y has elements*/
    {
        *head = y;
        return;
    }
    if((y==NULL))   /*if y is empty but x has elements*/
    {
        *head = x;
        return;
    }
    /*only other case remaining is that both list have elements*/
    while((x!=NULL)&&(y!=NULL))
    {
        if(*s==NULL)    /*adding first element to the unified list*/
        {
            *s = malloc(sizeof(struct node));
            z = *s;
        }
        else
        {
            z->next = malloc(sizeof(struct node));
            z = z->next;
        }

        if(x->data < y->data)
        {
            z->data = x->data;
            x = x->next;
        }
        else if(x->data > y->data)
        {
            z->data = y->data;
            y=y->next;
        }
        else if(x->data == y->data)
        {
            z->data = x->data;
            x = x->next;
            y = y->next;
        }
    }
    /*now if one list got exhausted and the other remains*/
    while(x!=NULL)
    {
        z->next = malloc(sizeof(struct node));
        z = z->next;
        z->data = x->data;
        x = x->next;
    }
    while(y!=NULL)
    {
        z->next = malloc(sizeof(struct node));
        z = z->next;
        z->data = y->data;
        y = y->next;
    }
    z->next = NULL;
    *head = *s;
}

/* Function to print nodes in a given linked list */
void printList(struct node *node)
{
  while(node!=NULL)
  {
   printf("%d ", node->data);
   node = node->next;
  }
}

/* Function to insert a node at the beginging of the linked list */
void push(struct node** head_ref, int new_data)
{
  /* allocate node */
  struct node* new_node =
            (struct node*) malloc(sizeof(struct node));

  /* put in the data  */
  new_node->data  = new_data;

  /* link the old list off the new node */
  new_node->next = (*head_ref);

  /* move the head to point to the new node */
  (*head_ref)    = new_node;
}

/* Drier program to test above functions*/
int main()
{
  /* Start with the empty list */
  struct node* res = NULL;
  struct node* p = NULL;

  /* Let us create a unsorted linked lists to test the functions
   Created lists shall be a: 2->3->20->5->10->15 */
  push(&p, 15);
  push(&p, 10);
  push(&p, 5);
  push(&p, 20);
  push(&p, 3);
  push(&p, 2);

  /* Sort the above created Linked List */
  mergesort(&p);

  printf("\n Sorted Linked List is: \n");
  printList(p);

  getchar();
  return 0;
}