C++ Program For Comparing Two Strings Represented As Linked Lists

Given two strings, represented as linked lists (every character is a node in a linked list). Write a function compare() that works similar to strcmp(), i.e., it returns 0 if both strings are the same, 1 if the first linked list is lexicographically greater, and -1 if the second string is lexicographically greater.
Examples:

Input: list1 = g->e->e->k->s->a
       list2 = g->e->e->k->s->b
Output: -1

Input: list1 = g->e->e->k->s->a
       list2 = g->e->e->k->s
Output: 1

Input: list1 = g->e->e->k->s
       list2 = g->e->e->k->s
Output: 0

C++

// C++ program to compare two strings
// represented as linked lists
#include<bits/stdc++.h>
using namespace std;

// Linked list Node
// structure
struct Node
{
    char c;
    struct Node *next;
};

// Function to create newNode
// in a linkedlist
Node* newNode(char c)
{
    Node *temp = new Node;
    temp->c = c;
    temp->next = NULL;
    return temp;
};

int compare(Node *list1,
            Node *list2)
{
    // Traverse both lists. Stop when
    // either end of a linked list is
    // reached or current characters
    // don't match
    while (list1 && list2 &&
           list1->c == list2->c)
    {
        list1 = list1->next;
        list2 = list2->next;
    }

    //  If both lists are not empty,
    // compare mismatching characters
    if (list1 && list2)
        return ((list1->c >
                 list2->c)? 1: -1);

    // If either of the two lists has
    // reached end
    if (list1 && !list2) return 1;
    if (list2 && !list1) return -1;

    // If none of the above conditions
    // is true, both lists have reached
    // end
    return 0;
}

// Driver code
int main()
{
    Node *list1 = newNode('g');
    list1->next = newNode('e');
    list1->next->next = newNode('e');
    list1->next->next->next =
           newNode('k');
    list1->next->next->next->next =
           newNode('s');
    list1->next->next->next->next->next =
           newNode('b');

    Node *list2 = newNode('g');
    list2->next = newNode('e');
    list2->next->next = newNode('e');
    list2->next->next->next =
           newNode('k');
    list2->next->next->next->next =
           newNode('s');
    list2->next->next->next->next->next =
           newNode('a');

    cout << compare(list1, list2);
    return 0;
}

Output:

1

Time Complexity: O(M + N), where M and N represents the length of the given two linked lists.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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