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]`.

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))

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