solved binary_tree_level_order_traversal #42
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oda
reviewed
Feb 15, 2025
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| if len(node_ids) > 0: | ||
| level_to_nodes.append(node_ids) |
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単に if node_ids: でもいいでしょう。
個人的には current_level_nodes = next_level_nodes の前に置きたいです。
current_level_nodes に None が入っているので、これいるんですね。よしあしですが、入るかもしれない前提で書いたほうが単純かもしれません。
| current_level_nodes = [root] | ||
| while current_level_nodes: | ||
| next_level_nodes = [] | ||
| node_ids = [] |
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問題文の単語に合わせて、node_vals としてもよいように感じました。他の単語に合わせて current_level_vals とかでもよいかもしれません。
| # self.right = right | ||
| class Solution: | ||
| def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: | ||
| level_to_nodes = [] |
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こちら、階層ごとの値が入っているというよりは、階層とそれに紐づく node が入っているように見えました。level_order_vals や all_level_vals が浮かびました。
TORUS0818
reviewed
Feb 24, 2025
| class Solution: | ||
| def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: | ||
| level_traversed_node_ids = [] | ||
| def append_by_depth(node: Optional[TreeNode], depth: int): |
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自分ならids -> vals、depth -> levelを使うと思います。
表現にバリエーションが出てしまうと、後々混乱の原因になるかもと思いました。
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問題: https://leetcode.com/problems/binary-tree-level-order-traversal/description/?envType=problem-list-v2&envId=xo2bgr0r