Print middle element of linked list using recursion in python

Python program for Print middle element of linked list using recursion. Here mentioned other language solution.

#  Python 3 program for
#  Find middle of singly linked list using recursion

#  Linked list node
class LinkNode :
    def __init__(self, data) :
        self.data = data
        self.next = None
class SingleLL :
    def __init__(self) :
        self.head = None
        self.tail = None
    #  Add new Node at end of linked list
    def addNode(self, data) :
        node = LinkNode(data)
        if (self.head == None) :
            self.head = node
        else :
            #  Append the node at last position
            self.tail.next = node
        self.tail = node
    #  Display linked list element
    def display(self) :
        if (self.head == None) :
            print("\n Empty linked list")
            return
        temp = self.head
        # iterating linked list elements
        while (temp != None) :
            print("",temp.data, end = " →")
            #  Visit to next node
            temp = temp.next
        print(" NULL")
    #  Find and print middle element
    def printMiddle(self, node,  mid) :
        if (node != None and 
            node.next != None and 
            node.next.next != None) :
            #  Recursively finding the middle node
            self.printMiddle(node.next.next, mid.next)
        else :
            #  When we get middle node
            print(" Middle Node :",mid.data)
    #  Handling the the request of 
    #  finding middle of given linked list
    def findMiddle(self) :
        if (self.head == None) :
            print(" Empty Linked List")
        elif (self.head.next == None) :
            #  When single node exist
            print(" Linked list contains only one node")
        else :
            print("\n Linked List ")
            self.display()
            self.printMiddle(self.head, self.head)
def main() :
    sll1 = SingleLL()
    sll2 = SingleLL()
    #  First linked list
    #  1 → 2 → 3 → 4 → -5 → 6 → 7 → NULL
    sll1.addNode(1)
    sll1.addNode(2)
    sll1.addNode(3)
    sll1.addNode(4)
    sll1.addNode(5)
    sll1.addNode(6)
    sll1.addNode(7)
    #  Second linked list
    #  4 → 9 → 7 → 3 → 8 → 6 → NULL
    sll2.addNode(4)
    sll2.addNode(9)
    sll2.addNode(7)
    sll2.addNode(3)
    sll2.addNode(8)
    sll2.addNode(6)
    #  Test A
    sll1.findMiddle()
    #  Test B
    sll2.findMiddle()


if __name__=="__main__":
    main()

Output

 Linked List
 1 → 2 → 3 → 4 → 5 → 6 → 7 → NULL
 Middle Node : 4

 Linked List
 4 → 9 → 7 → 3 → 8 → 6 → NULL
 Middle Node : 7