问链表的合并排序代码只对C语言中一半的元素进行排序

EN

struct Node { 
    void *data; 
    struct Node* next; 
}; 

int main() 
{ 

    struct Node* a = NULL; 
    struct Node* sorted = NULL;

    push(&a, "orange"); 
    push(&a, "banana"); 
    push(&a, "strawberry"); 
    push(&a, "apple"); 
    push(&a, "kiwi"); 
    push(&a, "grapes"); 

    sorted = MergeSort(a); 

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

    return 0; 
} 

/* sorts the linked list by changing next pointers (not data) */
struct Node* MergeSort(struct Node* headRef) 
{ 
    struct Node* head = headRef; 
    struct Node* a; 
    struct Node* b; 

    /* Base case -- length 0 or 1 */
    if ((head == NULL) || (head->next == NULL)) { 
        return headRef; 
    } 

    /* Split head into 'a' and 'b' sublists */
    FrontBackSplit(head, &a, &b); 

    /* Recursively sort the sublists */
    MergeSort(a); 
    MergeSort(b); 

    /* answer = merge the two sorted lists together */
    headRef = SortedMerge(a, b); 
    return headRef;
} 


struct Node* SortedMerge(struct Node* a, struct Node* b) 
{ 
    struct Node* result = NULL; 

    /* Base cases */
    if (a == NULL) 
        return (b); 
    else if (b == NULL) 
        return (a); 

    /* Pick either a or b, and recur */
    if (strcmp(a->data, b->data) <= 0) { 
        result = a; 
        result->next = SortedMerge(a->next, b); 
    } 
    else { 
        result = b; 
        result->next = SortedMerge(a, b->next); 
    } 
    return (result); 
} 

/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves, 
    and return the two lists using the reference parameters. 
    If the length is odd, the extra node should go in the front list. 
    Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct Node* source, 
                    struct Node** frontRef, struct Node** backRef) 
{ 
    struct Node* fast; 
    struct Node* slow; 
    slow = source; 
    fast = source->next; 

    /* Advance 'fast' two nodes, and advance 'slow' one node */
    while (fast != NULL) { 
        fast = fast->next; 
        if (fast != NULL) { 
            slow = slow->next; 
            fast = fast->next; 
        } 
    } 

    /* 'slow' is before the midpoint in the list, so split it in two 
    at that point. */
    *frontRef = source; 
    *backRef = slow->next; 
    slow->next = NULL; 
} 

Sorted Linked List is: 
grapes
kiwi
strawberry