链表的C++合并排序没有对前两个索引进行排序

#include <iostream>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

ListNode* merge(ListNode* l1, ListNode* l2) {
    if (l1 == nullptr) return l2;
    if (l2 == nullptr) return l1;
    
    if (l1->val < l2->val) {
        l1->next = merge(l1->next, l2);
        return l1;
    } else {
        l2->next = merge(l1, l2->next);
        return l2;
    }
}

ListNode* mergeSort(ListNode* head) {
    if (head == nullptr || head->next == nullptr) return head;
    
    ListNode* slow = head;
    ListNode* fast = head->next;
    
    while (fast != nullptr && fast->next != nullptr) {
        slow = slow->next;
        fast = fast->next->next;
    }
    
    ListNode* l1 = head;
    ListNode* l2 = slow->next;
    slow->next = nullptr;
    
    l1 = mergeSort(l1);
    l2 = mergeSort(l2);
    
    return merge(l1, l2);
}

void printList(ListNode* head) {
    ListNode* curr = head;
    while (curr != nullptr) {
        std::cout << curr->val << " ";
        curr = curr->next;
    }
    std::cout << std::endl;
}

int main() {
    ListNode* head = new ListNode(4);
    head->next = new ListNode(2);
    head->next->next = new ListNode(1);
    head->next->next->next = new ListNode(3);
    
    std::cout << "Original List: ";
    printList(head);
    
    head = mergeSort(head);
    
    std::cout << "Sorted List: ";
    printList(head);
    
    return 0;
}