diff --git a/Python Codes/Leetcode-question-Merge-Two-Sorted-Lists.py b/Python Codes/Leetcode-question-Merge-Two-Sorted-Lists.py
new file mode 100644
index 0000000..ae98c1b
--- /dev/null
+++ b/Python Codes/Leetcode-question-Merge-Two-Sorted-Lists.py	
@@ -0,0 +1,19 @@
+def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: 
+        p1=list1 #pointing to the head of list1
+        p2=list2 #pointing to the head of list2
+        dn=ListNode()  #made new node so that it will start to point the numbers of both lists in sorted order
+        p3=dn #pointing to the head of new node called 'dn'
+        while p1!=None and p2!=None:  #loop goes till head of either lists not reached to null node
+            if p1.val < p2.val:  #compairing data of both the heads
+                p3.next=p1 #the p3 will point the shortest head's data (here shortest is p1)
+                p1=p1.next #move the p1 to the next node (after get pointed by p3 in previous step)
+            else: 
+                p3.next=p2  #the p3 will point the shortest head's data (here shortest is p2)
+                p2=p2.next #move the p2 to the next node (after get pointed by p3 in previous step)
+            p3=p3.next  #move the p3 to the next node (as from above conditions p3 will be pointing to either head of list1 or list2)
+			
+        if p1 != None:  #if p2 reached to null but list1 nodes still remaining to be accessed 
+            p3.next=p1 #so p3 will point the remaining nodes of list1 
+        if p2 != None:  #if p1 reached to null but list1 nodes still remaining to be accessed 
+            p3.next=p2 #so p3 will point the remaining nodes of list2
+        return dn.next #the last node will be the head of merging sorted lists (which is asked to be returned)