**Objective:** Given two sorted linked lists, objective is to merge both the lists in sorted order.

**Input:** Two sorted linked list List a, List b.

Example:List a :->2->6->18Listb:->1->3->17->19Merged List: ->1->2->3->6->17->18->19

**Approach:**

**Without Recursion:**

- Create a new node say result
- Navigate through both the linked lists at the same time, starting from head
- Compare the first node values of both the linked lists
- Whichever is smaller, add it to the result node
- Move the head pointer of the linked list whose value was smaller
- Again compare the node values
- Keep doing until one list gets over
- Copy the rest of the nodes of unfinished list to the result

**With Recursion:**

- Base Case :

If List A gets finished , return List B.

If List B gets finished, return List A.

- Create a result node and initialize it as NULL
- Check which node (List A or List B) has a smaller value.
- Whichever is smaller, add it to the result node.
- Make recursive call and add the return node as result.next

result.next = recurrsionMerge(nA.next, nB)

**Complete Code for Both Methods:**

**Output**:

Method : with Recursion ->5->6->8 ->1->3->7->9 ->1->3->5->6->7->8->9 Method : without Recursion ->2->6->18 ->1->3->17->19 ->1->2->3->6->17->18->19