105.construct binary tree from preorder and inorder traversal#28
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105.construct binary tree from preorder and inorder traversal#28
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huyfififi
reviewed
Mar 26, 2026
| node.left = gather_descendants(node_position) | ||
| stack.append(node) | ||
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| return gather_descendants(-(10**30)) |
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この数字がどうやって決まっているのかわかりづらいので、Python にするのだったら float("-inf") とかの方が意図が伝わりやすいかもしれません。
| inorder_val_to_idx[val] = i | ||
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| def buildTree_w_index( | ||
| pre_left: int, in_left: int, num_children: int |
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私はこの書き方を検討しきれていませんでした。確かに、子の数がわかっているのでこの値に名前をつけて引数にするのが、とてもわかりやすく感じました。
colorbox
reviewed
Mar 26, 2026
| if num_children == 1: | ||
| return root | ||
| idx_root_inorder = inorder_val_to_idx[root.val] | ||
| num_left_children = idx_root_inorder - in_left |
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好みの範疇ですが、numよりはcountの方が個数ということを読み取りやすいと思いました。
colorbox
reviewed
Mar 26, 2026
| for i, val in enumerate(inorder): | ||
| inorder_val_to_idx[val] = i | ||
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| def buildTree_w_index( |
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メソッド名のwが何かを読み取れなかったので、略ではなく単語を使うことをお勧めします。
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withの略でしたが、わかりづらいのかもしれませんね。省略せずに書こうと思います。
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https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/