Only enqueue non-null children in binary tree level order traversal

[Copilot]

The level order traversal was appending n.left and n.right unconditionally, enqueueing None values and causing an extra iteration through a full level of nulls at the end.

Changes

Added null checks before appending children to the next level:

# Before
for n in nodes:
    if n:
        level.append(n.val)
        new_nodes.append(n.left)   # appends even if None
        new_nodes.append(n.right)  # appends even if None

# After  
for n in nodes:
    if n:
        level.append(n.val)
        if n.left:
            new_nodes.append(n.left)
        if n.right:
            new_nodes.append(n.right)

Eliminates redundant iteration and reduces memory overhead.

Original prompt

Problem

Based on this review feedback from PR #86:

new_nodes appends n.left/n.right even when they are None, which can enqueue a full level of None values and add an extra iteration that does no useful work. It's more efficient to only enqueue non-null children (or to build the next level list from actual children) to avoid the redundant pass and unnecessary memory use.

Current Code

In src/tree/binary_tree_level_order_traversal.py, lines 47-52:

new_nodes = []
for n in nodes:
    if n:
        level.append(n.val)
        new_nodes.append(n.left)
        new_nodes.append(n.right)

The issue is that n.left and n.right are appended to new_nodes unconditionally (they could be None). This means new_nodes can contain None values, leading to an extra unnecessary iteration through a list of all None values on the next pass.

Required Change

Add None checks before appending children to new_nodes. Only enqueue non-null children:

new_nodes = []
for n in nodes:
    if n:
        level.append(n.val)
        if n.left:
            new_nodes.append(n.left)
        if n.right:
            new_nodes.append(n.right)

This avoids enqueueing None values, preventing redundant passes and unnecessary memory use.

This pull request was created from Copilot chat.

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