Delete Middle of Linked List

PROBLEM :

Given a singly linked list, delete middle  of the linked list. For example, if given linked list is 1->2->3->4->5 then linked list should be modified to 1->2->4->5

If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.

If the input linked list is NULL, then it shoudl remain NULL.

If the input linked list has 1 node, then this node should be deleted and new head should be returned.

Input:
You have to complete the method which takes one argument: the head of the linked list. You should not read any input from stdin/console.
The struct Node has a data part which stores the data and a next pointer which points to the next element of the linked list.
There are multiple test cases. For each test case, this method will be called individually.

Output:
Your function should return head  of the modified linked list.  If linked list is empty then it should return NULL.

Constraints:
1 <=T<= 50
1 <=N<= 1000
1 <=value<= 1000

Example:
Input:
2
5
1 2 3 4 5
6
2 4 6 7 5 1

Output:
1 2 4 5
2 4 6 5 1

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SIMPLE c++ IMPLEMENTATION :
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/* Link list Node
struct Node
{
    int data;
    struct Node* next;
}; */

// Deletes middle of linked list and returns head of the modified list

struct Node* deleteMid(struct Node *head)
{
    struct Node *temp1,*temp2,*ptr ;
   
    if(head==NULL)
        return head ;
       
    if(head->next==NULL)
    {
        free(head) ;
        head=NULL ;
        return head ;
    }
   
    temp1=head ;
    temp2=head->next->next ;
   
    while(temp2!=NULL&&temp2->next!=NULL)
    {
        temp1=temp1->next ;
        temp2=temp2->next->next ;
    }
   
    ptr=temp1->next ;
    temp1->next=ptr->next ;
    free(ptr) ;
   
    return head ;
}

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