The short answer to this is that, Python is a pass-by-object-reference language, not pass-by-reference as implied in the question. It means that:

  1. result and result_tail are two variables that happen to point at the same value
  2. Mutation / Changing of the underlying value (result_tail.next = ListNode(1)) will affect the value shown by result
  3. However, assigning / pointing the variable result_tail to another value will NOT affect the value of result
  4. result_tail = result_tail.next is assigning the next node of the node that is currently assigned by the variable

The following is an visualization of the values that are assigned to the variables (r = result, rt = result_tail):

result = ListNode(0)
#r
#0 -> None

result_tail = result
#r
#0 -> None
#rt

result_tail.next = ListNode(1)
#r
#0 -> 1 -> None
#rt

result_tail = result_tail.next
#r
#0 -> 1 -> None
#     rt

result_tail.next = ListNode(2)
#r
#0 -> 1 -> 2 -> None
#     rt

result_tail = result_tail.next
#r
#0 -> 1 -> 2 -> None
#          rt

References for additional reading:

  • An article explaining the Python pass-by-object reference style in detail https://robertheaton.com/2014/02/09/pythons-pass-by-object-reference-as-explained-by-philip-k-dick/
  • An answer explaining Python's pass-by-object reference style https://stackoverflow.com/a/33066581/12295149
  • Question asking on Python's object reference style Understanding Python's call-by-object style of passing function arguments
Answer from Joseph on Stack Overflow
๐ŸŒ
GeeksforGeeks
geeksforgeeks.org โ€บ python โ€บ python-linked-list
Python Linked List - GeeksforGeeks
December 11, 2025 - class Node: def __init__(self, data): self.data = data self.next = None # Create nodes node1 = Node(15) node2 = Node(3) node3 = Node(17) node4 = Node(90) # Link nodes node1.next = node2 node2.next = node3 node3.next = node4 head = node1 # Head points to the first node # Traverse and print the linked list current = head while current: print(current.data, end=" -> ") current = current.next print("None")
Top answer
1 of 4
27

The short answer to this is that, Python is a pass-by-object-reference language, not pass-by-reference as implied in the question. It means that:

  1. result and result_tail are two variables that happen to point at the same value
  2. Mutation / Changing of the underlying value (result_tail.next = ListNode(1)) will affect the value shown by result
  3. However, assigning / pointing the variable result_tail to another value will NOT affect the value of result
  4. result_tail = result_tail.next is assigning the next node of the node that is currently assigned by the variable

The following is an visualization of the values that are assigned to the variables (r = result, rt = result_tail):

result = ListNode(0)
#r
#0 -> None

result_tail = result
#r
#0 -> None
#rt

result_tail.next = ListNode(1)
#r
#0 -> 1 -> None
#rt

result_tail = result_tail.next
#r
#0 -> 1 -> None
#     rt

result_tail.next = ListNode(2)
#r
#0 -> 1 -> 2 -> None
#     rt

result_tail = result_tail.next
#r
#0 -> 1 -> 2 -> None
#          rt

References for additional reading:

  • An article explaining the Python pass-by-object reference style in detail https://robertheaton.com/2014/02/09/pythons-pass-by-object-reference-as-explained-by-philip-k-dick/
  • An answer explaining Python's pass-by-object reference style https://stackoverflow.com/a/33066581/12295149
  • Question asking on Python's object reference style Understanding Python's call-by-object style of passing function arguments
2 of 4
16

For those reading this in the future: I wanted to debug linked list problems on a local environment so here is what I did.

  1. Modified the Leetcode code for ListNode by including the dunder "repr" method. This is for when you want to print a ListNode to see what its value and next node(s).
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

    def __repr__(self):
        return "ListNode(val=" + str(self.val) + ", next={" + str(self.next) + "})"
  1. Next, I made a recursive function that makes a nested ListNode when you pass in a list. This is so you can test your methods by passing in lists (instead of having to manually make a confusing looking ListNode yourself.
def list_to_LL(arr):
    if len(arr) < 1:
        return None

    if len(arr) == 1:
        return ListNode(arr[0])
    return ListNode(arr[0], next=list_to_LL(arr[1:]))
  1. Here is an example that tests my answer for the "reverseList" problem:
def reverseList(head: ListNode) -> ListNode:
    prev = None
    while head:
        next_node = head.next
        head.next = prev
        prev = head
        head = next_node

    return prev


# test cases
t1 = list_to_LL([1, 2, 3, 4, 5])  #ListNode(val=1, next={ListNode(val=2, next={ListNode(val=3, next={ListNode(val=4, next={ListNode(val=5, next={None})})})})})
t2 = list_to_LL([1, 2])  #ListNode(val=1, next={ListNode(val=2, next={None})})
t3 = list_to_LL([])

# answers
print(reverseList(t1))
print(reverseList(t2))
print(reverseList(t3))
๐ŸŒ
Stack Abuse
stackabuse.com โ€บ python-linked-lists
Python Linked Lists
August 25, 2023 - To have a data structure we can work with, we define a node. We'll implement a node as a class named ListNode. The class contains the definition to create an object instance, in this case, with two variables - data to keep the node value, and next to store the reference to the next node in the list.
๐ŸŒ
CodeSignal
codesignal.com โ€บ learn โ€บ courses โ€บ getting-deep-into-complex-algorithms-for-interviews-with-python โ€บ lessons โ€บ linked-list-operations-in-python
Linked List Operations in Python
class ListNode: def __init__(self, value=0, next=None): self.value = value # Holds the value or data of the node self.next = next # Points to the next node in the linked list; default is None # Initialization of linked list head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))
๐ŸŒ
W3Schools
w3schools.com โ€บ python โ€บ python_dsa_linkedlists.asp
Linked Lists with Python
Finding the lowest value in a singly linked list in Python: class Node: def __init__(self, data): self.data = data self.next = None def findLowestValue(head): minValue = head.data currentNode = head.next while currentNode: if currentNode.data < minValue: minValue = currentNode.data currentNode = currentNode.next return minValue node1 = Node(7) node2 = Node(11) node3 = Node(3) node4 = Node(2) node5 = Node(9) node1.next = node2 node2.next = node3 node3.next = node4 node4.next = node5 print("The lowest value in the linked list is:", findLowestValue(node1)) Run Example ยป
๐ŸŒ
Reddit
reddit.com โ€บ r/learnpython โ€บ [deleted by user]
[deleted by user] : r/learnpython
December 5, 2023 - # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next
๐ŸŒ
Statistics Globe
statisticsglobe.com โ€บ home โ€บ python programming language for statistics & data science โ€บ what is a list node in python? (2 examples)
What is a List Node in Python? (2 Examples) | Linked List Structure
May 15, 2023 - class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next # instantiate the nodes node1 = ListNode("The") node2 = ListNode("boy") node3 = ListNode("is") node4 = ListNode("tall") # link the nodes node1.next = node2 node2.next = node3 node3.next = node4 # traverse the linked list and print each node's value current_node = node1 while current_node is not None: print(current_node.val) current_node = current_node.next # The # boy # is # tall
Find elsewhere
๐ŸŒ
CodeRivers
coderivers.org โ€บ blog โ€บ listnode-python
Understanding and Using ListNode in Python - CodeRivers
July 21, 2026 - A ListNode is a node in a linked list. It can be thought of as a container that holds some data and a reference (or link) to the next node in the list. In Python, we can represent a ListNode as a class.
๐ŸŒ
Real Python
realpython.com โ€บ linked-lists-python
Linked Lists in Python: An Introduction โ€“ Real Python
June 24, 2026 - In the above class definition, you can see the two main elements of every single node: data and next. You can also add a __repr__ to both classes to have a more helpful representation of the objects:
๐ŸŒ
CodeRivers
coderivers.org โ€บ blog โ€บ how-to-create-listnode-in-python
Creating ListNodes in Python: A Comprehensive Guide - CodeRivers
February 22, 2026 - In this code: - The __init__ method is the constructor of the ListNode class. - The val parameter represents the value of the node, and it has a default value of 0.
๐ŸŒ
GitHub
github.com โ€บ mcclee โ€บ Leetcode-python-Listnode
GitHub - mcclee/Leetcode-python-Listnode: A python class to convert list to Listnode ยท GitHub
A python class to convert list to Listnode Usage: from ListToListnode import FuckListnode list1 = [1, 2, 3, 4] f = FuckListnode() listnode = f.returnNode(list1)
Author: mcclee
๐ŸŒ
Python Forum
python-forum.io โ€บ thread-31071.html
How to create a linked list and call it?
I want to create a linked list and insert dummy data to verify it. Please see my code. class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next a = ListNode(2) a.next = Li
๐ŸŒ
Reddit
reddit.com โ€บ r/learnpython โ€บ help me out with listnode
r/learnpython on Reddit: Help me out with ListNode
July 10, 2025 -

Hello all, I completed my 12th this may( high school graduate ) going to attend Engineering classes from next month. So I decided to start LeetCode question. Till now I have completed about 13 questions which includes 9 easy ones, 3 medium ones and 1 hard question( in python language ) with whatever was thought to me in my school, but recently I see many questions in from ***ListNode***, but searching in youtube doesn't shows anything about ListNode but only about Linked list. So kindly suggest me or provide the resources to learn more about it.

Thank you!

๐ŸŒ
TutorialsPoint
tutorialspoint.com โ€บ python_data_structure โ€บ python_linked_lists.htm
Python - Linked Lists
This involves changing the pointer of a specific node to point to the new node. That is possible by passing in both the new node and the existing node after which the new node will be inserted.
๐ŸŒ
DataCamp
datacamp.com โ€บ tutorial โ€บ python-linked-lists
Python Linked Lists: Tutorial With Examples | DataCamp
June 2, 2026 - Next, we need to create the linked list class. This will encapsulate all the operations for managing the nodes, such as insertion and removal.
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Towards Data Science
towardsdatascience.com โ€บ home โ€บ data science โ€บ how to implement a linked list in python
How to Implement a Linked List in Python | Towards Data Science
December 20, 2021 - The full code containing the three classes we created as part of today's tutorial is given below as a GitHub Gist. In today's guide we discussed about one of the most fundamental data structures, namely Linked Lists. Given that Python's standard library does not contain any implementation of this specific data structure, we explored how one can implement a user-defined Linked List class from scratch.
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Built In
builtin.com โ€บ data-science โ€บ python-linked-list
An Introduction to Python Linked List and How to Create One
Thus, we have implemented the main functionality of a singly linked list in Python. All that remains is to put this all together, which can be done as follows: Class Node(object): def __init__(self, val): self.val = val self.next = None def get_data(self): return self.val def set_data(self, val): self.val = val def get_next(self): return self.next def set_next(self, next): self.next = next Class LinkedList(object): def __init__(self, head = None): self.head = head self.count = 0 def insert(self, data): """ Create a new node at the Head of the Linked List """ #create a new node to hold the data
๐ŸŒ
Python.org
discuss.python.org โ€บ python help
An easy leetcode question - Python Help - Discussions on Python.org
February 23, 2022 - Moreover, Leetcode provides a list, not a linked list. Do I have to write my own linked list? This is from the original question: # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None I do not have the soluti...
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freeCodeCamp
freecodecamp.org โ€บ news โ€บ introduction-to-linked-lists-in-python
Linked Lists in Python โ€“ Explained with Examples
September 22, 2022 - class LinkedList: def __init__(self,head=None): self.head = head def append(self, new_node): current = self.head if current: while current.next: current = current.next current.next = new_node else: self.head = new_node
Top answer
1 of 4
7

Summary

I won't dwell on what has already been cited by users toolic and J_H, so I just have a few comments:

Type Hinting

I would suggest that you include type hinting, especially if your functions do not contain docstrings that describe the type of arguments being passed to functions (J_H has suggested this, so pardon if this is too repetitive).

Be More Tolerant of Errors in User Input

If the user does not enter a valid integer in function run_and_add, you essentially quit. You should instead put out the prompt again and give the user as many chances needed to enter valid input. The user can always terminate by entering Ctrl-C if they get stuck.

Strive for Encapsulation and Reusability

I can't stress too strongly that your code is crying out for you to create a LinkedList abstract data type that abstracts the notion of a linked list while encapsulating the actual implementation. To that end, I would use attribute names that begin with '_' where appropriate to suggest that they are "private" and not to be either updated nor depended on existing in the future (such as the next instance attribute of the ListNode class.

The following classes are just one possibility. Note:

  1. There is no print method implemented since printing the entire list is trivial given that the class implements the iterator protocol. Besides, what if you wanted to print to a file? Then a print method would require one or more additional arguments.
  2. The client never explicitly creates ListNode instances.
  3. The linked list keeps explicit track of the final (last) node in the list to provide efficient appending of a node or an entire linked list to the end.

I can envision your using this as a starting point and potentially adding other methods (for example, __eq__ methods to compare nodes and linked lists).

"""A module for creating and manipulating linked lists."""

from abc import ABC, abstractmethod
from typing import TypeVar, Any

LinkedListInstance = TypeVar('LinedListInstance', bound='LinkedList')

class NodeType(ABC):
    @property
    @abstractmethod
    def val(self):
        pass

    @val.setter
    @abstractmethod
    def val(self, val):
        pass

class LinkedList:

    class _ListNode(NodeType):
        """Initialize a new node with some value, val."""

        def __init__(self, val):
            self._val = val
            self._next = None

        @property
        def val(self):
            return self._val

        @val.setter
        def val(self, val):
            self._val = val

        def __repr__(self):
            return f'_ListNode({repr(self._val)})'

        def __str__(self):
            return str(self._val)

    def __init__(self):
        """Create a new, empty linked list."""

        self._head = None
        self._tail = None

    def append_node(self, val: Any) -> LinkedListInstance:
        """Append a new node to the list initialized with val."""

        new_node = LinkedList._ListNode(val)
        if self._head is None:
            self._head = new_node
        else:
            self._tail._next = new_node
        self._tail = new_node

        return self

    def insert_node(self, at_node: NodeType, val: Any) -> LinkedListInstance:
        """Create and insert a new node after the specified at_node node initialized
        with val."""

        if at_node is self._tail:  # special case
            return self.append_node(val)
        node_to_insert = LinkedList._ListNode(val)
        node_to_insert._next = at_node._next
        at_node._next = node_to_insert

        return self

    def append_list(self, linked_list: LinkedListInstance) -> LinkedListInstance:
        """Append a linked list to the current list."""

        if self._head is None:
            self._head = linked_list._head
        else:
            self._tail._next = self._head
        self._tail = linked_list._tail

        return self

    def __iter__(self) -> NodeType:
        """Iterate the list."""

        current = self._head
        while current is not None:
            yield current
            current = current._next

if __name__ == '__main__':
    def insert_node_at_position(linked_list: LinkedList, position: int) -> None:
        for counter, current_node in enumerate(linked_list, start=1):
            print(f"Node at position {counter}: {current_node}")
            if counter == position:
                while True:
                    try:
                        number = int(input("Please insert an Integer: "))
                    except ValueError:
                        print("Not an Integer")
                    else:
                        break
                linked_list.insert_node(current_node, number)
                print("Node added at position:", position)

        print("Updated linked list:")
        for node in linked_list:
            print(node)

    linked_list = LinkedList().append_node(1).append_node(2).append_node(3)
    insert_node_at_position(linked_list, 2)
2 of 4
6

names

class ListNode:

This is a perfectly fine identifier, as-is.

There's no adjacent code that uses other node types. Consider shortening to just Node.

design of Public API

OO

def print_linked_list(head):
...
def add_node(prev_node, node_to_add):
...

These are somewhat unexpected signatures, the sort of thing I might expect in Fortran code. ListNode turned out to be just a very brief @dataclass, with no OO aspect to it. Given a ListNode, we find no methods to call on it for list operations. This works, but makes it a little harder for developers and maintenance engineers to discover your API. For example if I hit a breakpoint() I cannot p dir(node) to find plausible things I might do with a node -- I instead have to scour the codebase for such operations.

Also, your signatures lack ListNode type annotations, so I can't just grep for that or use type-aware IDE features to narrow my search.

I propose some more natural implementations.

    def print_linked_list(self):
        head = self
        while head:
            print(head.val)
            head = head.next

    def add_node(self, node_to_add):
        assert node_to_add.next is None
        node_to_add.next = self.next
        self.next = node_to_add

Consider renaming these to simply .print() and .insert().

interactive input vs parameter

(I am paraphrasing, renaming the vague number to new_val.)

def run_and_add(head, position):
                ...
                new_val = int(input("Please insert an Integer: "))

Prefer to place calls of input() further up in the call stack, such as within def main():, and pass in such a value as a parameter:

def run_and_add(head, position, new_val):

main guard

On which topic, you don't have a main() function, and you really need one. Why? So you or some maintenance engineer can safely import linkedlist when exercising your functions in a test suite. Also, it's convenient to ensure that local variables like first (which are not part of your exported Public API) will disappear when they go out of scope. That way such identifiers won't pollute the module namespace.

def main():
    first = ListNode(1)
    first.next = ListNode(2)
    first.next.next = ListNode(3)
    run_and_add(first, 2)

if __name__ == '__main__':
    main()

single responsibility

run_and_add() is an awkward identifier, suggesting that instead of one we're doing two things. Also I find "run" less than clear.

Consider making caller responsible for passing in an already-created node, and then this could be a simple insert_at_position(head, position, new_node) function.