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mamo3gr
reviewed
Feb 7, 2026
| def merge_node(child1, child2): | ||
| if child1 is None and child2 is None: | ||
| return | ||
| if child1 and child2: |
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[fyi]
Googleのスタイルガイドでは、(not) None 判定にはimplicit falseを使わないことが推奨されています。
https://google.github.io/styleguide/pyguide.html#2144-decision
Always use if foo is None: (or is not None) to check for a None value.
mamo3gr
reviewed
Feb 7, 2026
Comment on lines
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| if node1 is None: | ||
| node1 = TreeNode() | ||
| if node2 is None: | ||
| node2 = TreeNode() | ||
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| for side in ("left", "right"): | ||
| child1 = getattr(node1, side) | ||
| child2 = getattr(node2, side) |
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node1 がNoneのときでもleft(right)を引きたい、というのにTreeNodeを使うのは冗長に感じました。
Suggested change
| if node1 is None: | |
| node1 = TreeNode() | |
| if node2 is None: | |
| node2 = TreeNode() | |
| for side in ("left", "right"): | |
| child1 = getattr(node1, side) | |
| child2 = getattr(node2, side) | |
| for side in ("left", "right"): | |
| child1 = getattr(node1, side) if node1 is not None else None | |
| child2 = getattr(node2, side) if node2 is not None else None |
mamo3gr
reviewed
Feb 7, 2026
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| merged_child = merge_node(child1, child2) | ||
| if merged_child: | ||
| setattr(frontier_node, side, merged_child) |
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個人的な感覚としては、getattr や setattr は何か特別な事情があるときの手段で、似たようなコードの重複を減らすために使うのは過剰に感じました。
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解く問題
Merge Two Binary Trees
次に解く問題
Convert Sorted Array To Binary Search Tree