Add 1 to a number represented as linked list

PROBLEM :

A number (n) is represented in Linked List such that each digit corresponds to a node in linked list. Add 1 to it.

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.. There are multiple test cases. For each test case, this method will be called individually

Output:
Your function should return pointer to head of the modified list.

Constraints:
1 <=T<= 50
1 <=n<= 10000

Example:

Input:
4
456
123
999
1879

Output:
457
124
1000
1880

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

// Should rearrange given linked list such that all even
// positioned Nodes are before odd positioned.
// Returns new head of linked List.

struct Node* reverse(struct Node *) ;

Node *addOne(Node *head)
{
   if(head==NULL)
        return head ;
     
    int sum,rem ;
    head=reverse(head) ;
 
    Node *ptr, *temp=head ;
    rem=1 ;
 
    while(temp!=NULL)
    {
        sum=temp->data+rem ;
        temp->data=sum%10 ;
        rem=sum/10 ;
     
        temp=temp->next ;
    }
 
    if(rem!=0)
    {
        temp=head ;
        while(temp->next!=NULL)
            temp=temp->next ;
         
        ptr=(Node*)malloc(sizeof(Node)) ;
        ptr->data=rem ;
        ptr->next=NULL ;
        temp->next=ptr ;
    }
 
    head=reverse(head) ;
    return head ;
}

struct Node* reverse(struct Node *head)
{
    if(head==NULL)
    return head ;
 
    if(head->next==NULL)
    return head;
 
    struct Node *one,*two,*third ;
    one=head ;
    two=one->next ;
    third=two->next ;
 
    one->next=NULL ;
    two->next=one ;
 
    while(third!=NULL)
    {
        one=two ;
        two=third ;
        third=third->next ;
     
        two->next=one ;
    }
    return two ;
}

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SIMPLE c++ IMPLEMENTATION :(Better solution - without reversing, ie in single traversal)
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/* Node structure
struct Node
{
    int data;
    struct Node* next;
}; */

// Should rearrange given linked list such that all even
// positioned Nodes are before odd positioned.
// Returns new head of linked List.

Node* ADDone(Node *,int *) ;
Node *addOne(Node *head)
{
   if(head==NULL)
        return NULL ;
    int carry=0 ;
    head=ADDone(head,&carry) ;
   
    if(carry!=0)
    {
        Node *temp=(Node*)malloc(sizeof(Node)) ;
        temp->data=carry ;
        temp->next=head ;
        head=temp ;
    }
   
    return head ;
}

Node* ADDone(Node *head,int *carry)
{
    if(head==NULL)
        return NULL ;
       
    Node *prev ;
    prev=ADDone(head->next,&(*carry)) ;
   
    if(prev==NULL)
    {
        head->data=head->data+1 ;
        (*carry)=head->data/10 ;
        head->data=head->data%10 ;
    }
    else
    {
        head->data=head->data+(*carry) ;
        (*carry)=head->data/10 ;
        head->data=head->data%10 ;
    }
   
    head->next=prev ;
    return head ;
}

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