102. Binary Tree Level Order Traversal #26
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huyfififi
reviewed
Aug 10, 2025
| result = [] | ||
| nodes = [root] | ||
| while nodes: | ||
| result.append([node.val for node in nodes]) |
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個人的には、resultに先に追加してもfor loop内で追加していってもコードの読みやすさにあまり違いを感じなかったので、nodesを2回走査していることの方が少し気になりましたが、他の方々の感覚も気になるところですね 👀
（制約を見るに、一つの階層にあるNodeの数は小さく、2回走っても大した問題ではないですが。。）
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横から失礼します。nodes には同じレベルのノードが保存されていると思うので、好みかもしれませんが私も下の for ループ内で一緒に処理をしたいと思いました。
| class Solution: | ||
| def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: | ||
| if root is None: | ||
| return [] |
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step1 では、ここで 空の list で初期化された result を返されていましたが、自分もここでは [] を返すのが好みです。
tokuhirat
reviewed
Aug 11, 2025
| next_level_nodes.append(node.right) | ||
| nodes = next_level_nodes | ||
| return result | ||
| ``` |
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問題：102. Binary Tree Level Order Traversal
次の問題：103. Binary Tree Zigzag Level Order Traversal