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いいと思います。inorder順に構築する方法もあるみたいです |
olsen-blue
reviewed
Mar 2, 2025
| class Solution: | ||
| def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: | ||
| def index_range_to_BST(left: int, right: int) -> Optional[TreeNode]: | ||
| if not left < right: |
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僕はnot使わず書いてましたが、not 使うこちらの方が直感に近いかもですね。
(自然言語で表現すると)区間が存在しないとなるので、こちらの方がすんなりくる気がしました。
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ありがとうございます。二分探索もそうですが、これとかも、区間の取り方と成立不成立によって'='がついたりつかなかったりしてややこしいので、notを入れて統一するのが好みかもしれません
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わかります。探索アルゴリズムのグリッド範囲外判定とかも not でやりたい気持ちです。
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ここらへんは趣味かと思います。私は、left >= right と左が大きいときには否定的な意味としています。
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@SuperHotDogCat コメントありがとうございます。 |
hroc135
reviewed
Mar 8, 2025
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| middle = (left + right) // 2 | ||
| left = index_range_to_BST(left, middle) | ||
| right = index_range_to_BST(middle + 1, right) |
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下でコメントされていますが、int型の引数leftをTreeNodeとしても用いるのはあまり良くないと思います。
fuga-98
reviewed
Mar 17, 2025
| ```python | ||
| class Solution: | ||
| def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: | ||
| def index_range_to_BST(left: int, right: int) -> Optional[TreeNode]: |
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https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/