链表的归并排序错误

Question

我正试图在链表排序上做合并排序实践，我无法找出代码不起作用的原因，我的编译器没有给出任何有用的错误信息。就不能有人帮我指出来吗?谢谢!

    #include <stdio.h>
    #include <stdlib.h>
    #include <iostream>
    using namespace std;
    struct listnode { struct listnode * next; int key; };
    //function prototypes
    void seperate(struct listnode * node, struct listnode ** front, struct listnode ** back);
    struct listnode * merge(struct listnode * node_1, struct listnode * node_2);
    //merge sorted seperated linked list
    void mergeSort(struct listnode ** node)
    {
    struct listnode * head = * node; struct listnode * node_1; struct listnode * node_2;
    if ((head == NULL) || (head->next == NULL)) { return; }
    seperate(head, &node_1, &node_2);
    mergeSort(&node_1); mergeSort(&node_2);
    * node = merge(node_1, node_2);
}
//sort sepearted linked list
struct listnode * merge(struct listnode * node_1, listnode * node_2)
{
    struct listnode * return_result = NULL;
    if (node_1 == NULL) { return (node_2); }
    else if (node_2 = NULL) { return (node_1); }
    if (node_1->key <= node_2->key)
    {
        return_result = node_1; return_result->next = merge(node_1->next, node_2);
    }
    else { return_result = node_2; return_result->next = merge(node_1, node_2->next); }
    return return_result;
}
//function to seperate linked list 
void seperate(struct listnode * node, struct listnode ** front, struct listnode ** back)
{
    struct listnode * fast; struct listnode * slow;
    if (node == NULL || node->next == NULL) { *front = node; * back = NULL; }
    else
    {
        slow = node; fast = node->next;
        while (fast != NULL)
        {
            fast = fast->next;
            if (fast != NULL) { slow = slow->next; fast = fast->next; }
        }
        * front = node; * back = slow->next; slow->next = NULL;
    }

}// merge sort of linked list completed

 //test functions to push and print the linked list  
void push(struct listnode ** head, int data)
{
    struct listnode * added_node = (struct listnode * )malloc(sizeof(struct listnode));
    added_node->key = data;
    added_node->next = (*head);
    (*head) = added_node;
}
void printList(struct listnode * node)
{
    while (node != NULL)
    {
        cout << node->key;
        node = node->next;
    }
}
int main()
{
    struct listnode * node1 = NULL;
    push(&node1, 3); push(&node1, 30); push(&node1, 23); push(&node1, 1); push(&node1, 0); push(&node1, 9);
    mergeSort(&node1);
    cout << endl;
    printList(node1);
    return 0;
}

Answer 1

if (node_1 == NULL) { return (node_2); }
else if (node_2 = NULL) { return (node_1); }
/// -----------^ this is an assignment here

我将向尤达介绍你

if (NULL == node_1) { return (node_2); }
else if (NULL=node_2) { return (node_1); }
/// ---------^ still an assignment, 
//   but try it out & C what the compiler has to say