We will use the same way of addition as we use to do during elementary school, we will keep track of carry using a variable car and traverse digit by digit sum starting from the head of two linked list which contains least significant digit to most significant digits.

Pseudo Code:

Node sum(Node a,Node b){
	//Let Temp1 and temp2 be some Nodes
	int carry=0    // initilise carry as 0
	// temp2 is used as dummy for adding new node and 
	// temp1 will be use as the header for the 
	// sum value linked list
	temp2=temp1    
	int x,y        //used for storing nodes values
	while(a!=NULL or b!=NULL)
	{
		if(a!=NULL)
		{x=a->data,a=a->next;}
		else
		x=0
		if(b!=NULL)
		{y=b->data,b=b->next;}
		else
		y=0
		int sum=x+y+carry
		temp2->next=new Node(sum%10)   // store the digit(0-9)
		carry=sum/10                   // sum/10 gives carry
		temp2=temp2->next
	}
	if(carry!=0)      // if carry is not zero then add it at last.
		temp2->next=new Node(carry)
	return temp1->next // return the new Node head
}

Time complexity for above process is O(max(len(N1,N2))

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

struct Node //declare Node
    {
    Node* next;
    int data;
    Node(int data)
    {
        this->data = data; // node value
        this->next = NULL; // node pointer
    }
};

Node* addNode(Node* head, int value)
{
    // declare two nodes temp and p
    Node *temp, *p;
    // createNode will return a new node with
    // data = value and next pointing to NULL.
    temp = new Node(value);
    if (head == NULL) {
        head = temp; // when linked list is empty
    }
    else {
        p = head; // assign head to p
        while (p->next != NULL) {
            // traverse the list until p is the last node.
            // The last node always points to NULL.
            p = p->next;
        }
        // Point the previous last node to the
        // new node created.
        p->next = temp;
    }
    return head;
}

Node* sum(Node* a, Node* b)
{
    Node *temp1, *temp2; // initilise two Nodes
    temp1 = new Node(0);
    temp2 = temp1;
    int x, y;
    int carry = 0; // initially carry is zero
    while (a != NULL or b != NULL) {
        // if node a is NOT NULL then x=a->data
        if (a != NULL) {
            x = a->data;
            a = a->next;
        }
        else
            x = 0;
        if (b != NULL) {
            // if node a is NOT NULL then x=a->data
            y = b->data;
            b = b->next;
        }
        else
            y = 0;
        int sum1 = x + y + carry;
        temp2->next = new Node(sum1 % 10);
        carry = sum1 / 10;
        temp2 = temp2->next;
    }
    if (carry) // if carry is not zero
        temp2->next = new Node(carry);
    // return head pointer of the linked
    // list after summation.
    return temp1->next;
}

// function to print linked list
void print(Node* head)
{
    while (head != NULL) {
        if (head->next != NULL)
            cout << head->data << "->";
        else
            cout << head->data;
        head = head->next;
    }
}

int main()
{
    ll t;
    cout << "Enter number of test cases: ";
    cin >> t;

    while (t--) {
        cout << "Enter size of node N1: ";
        ll n1, n2;
        cin >> n1;

        Node *a, *b;
        ll x;

        cout << "Enter its elements: ";
        for (ll i = 0; i < n1; i++) {
            cin >> x;
            if (i == 0) {
                a = new Node(x);
            }
            else
                a = addNode(a, x);
        }

        cout << "Enter size of node N2: ";
        cin >> n2;

        cout << "Enter its elements: ";
        for (ll i = 0; i < n2; i++) {
            cin >> x;
            if (i == 0) {
                b = new Node(x);
            }
            else
                b = addNode(b, x);
        }

        cout << "The final sum is: ";
        print(sum(a, b));
        cout << "\n";
    }
    return 0;
}

Output

Enter number of test cases: 3
Enter size of node N1: 3
Enter its elements: 2 3 4
Enter size of node N2: 3
Enter its elements: 8 1 7
The final sum is: 0->5->1->1
Enter size of node N1: 3
Enter its elements: 2 1 3
Enter size of node N2: 4
Enter its elements: 1 2 3 4
The final sum is: 3->3->6->4
Enter size of node N1: 2
Enter its elements: 1 2
Enter size of node N2: 2
Enter its elements: 2 4
The final sum is: 3->6