# Add Two Numbers

You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

**Example 1:**

**Input:** l1 = [2,4,3], l2 = [5,6,4]

**Output:** [7,0,8]

**Explanation:** 342 + 465 = 807.

**Example 2:**

**Input:** l1 = [0], l2 = [0]

**Output:** [0]

**Example 3:**

**Input:** l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]

**Output:** [8,9,9,9,0,0,0,1]

**Constraints:**

- The number of nodes in each linked list is in the range
`[1, 100]`

. `0 <= Node.val <= 9`

- It is guaranteed that the list represents a number that does not have leading zeros.

Approach 1:

Recursive Solution

We go on adding both of the node values and the carry in a recursive manner , once we find any of the linked list is exhausted then the value assumed as 0 . Also we link each of the node to its previous node at the end of the recursion

`class Solution(object):`

def addTwoNumbers(self, l1, l2,carry=0):

#print(l1.val,l2.val,bool(l1),bool(l2))

val1,val2=0,0

val1=l1.val if l1 else 0

val2=l2.val if l2 else 0

carry,quotient=divmod(val1+val2+carry,10)

node=ListNode(quotient)

l1= l1.next if l1 else l1

l2= l2.next if l2 else l2

if l1 or l2 or carry:

node.next=self.addTwoNumbers(l1,l2,carry)

return node

Time: O(max(m,n)) m,n -> size of first and second linked list

Space: O(max(m,n))

Approach 2:

Same can be solved using the Iterative solution.

`class Solution(object):`

def addTwoNumbers(self, l1, l2):

carry=0

temp=ListNode(0)

node_=temp

while l1 or l2:

if l1 ==None:

one=0

else:

one=l1.val

if l2==None:

two=0

else:

two=l2.val

inter=ListNode((one+two+carry)%10)

temp.next=inter

temp=inter

carry=(one+two+carry)//10

if l1:

l1=l1.next

if l2:

l2=l2.next

if carry!=0:

temp.next=ListNode(carry)

return node_.next

Time: O(max(m,n)) m,n -> size of first and second linked list

Space: O(max(m,n))