Node class and left child value

[Rajeev Bagra]

Node class and left child value
byu/DigitalSplendid inlearnpython

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byu/DigitalSplendid from discussion
inlearnpython

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byu/DigitalSplendid from discussion
inlearnpython

Great question!

You’re asking: If the Node class’s constructor has a default argument like left_child=None, are we locking its value to None? And then, how can it later be checked with something like isinstance(left_child, Node)?

Let’s break it down.

1. Default Argument Doesn’t Lock the Value

The default argument left_child=None means:

If no value is passed during initialization, use None as the default.

But if a value is passed, that value overrides the default.

Example:

class Node:
    def __init__(self, value, left_child=None, right_child=None):
        self.value = value
        self.left = left_child
        self.right = right_child

        if left_child is not None:
            print(f"Left child is: {left_child.value}")

Then this will work just fine:

a = Node(1)
b = Node(2, left_child=a)

In this case:

• For a, no left or right child is passed → both default to None.
• For b, left_child=a is passed → left_child is now a Node object.

2. Using isinstance(left_child, Node)

You might see this used when validating the input:

if isinstance(left_child, Node):
    # safe to assume it's a Node object

This works because:

• If left_child was not passed, its value is None → isinstance(None, Node) returns False.
• If it was passed as a Node object → isinstance(left_child, Node) returns True.

So no conflict exists.

---

Summary

• left_child=None sets a default, not a permanent lock.
• If a Node object is passed, that value replaces the default.
• You can safely check if it’s a Node using isinstance().

Let me know if you want this visually illustrated or explained using your Canva design.

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