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Yuto729
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arahi10
reviewed
Mar 11, 2026
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| 再帰的にサブツリーを構成していけば良いと考えた。サブツリーの具体的な構成方法について | ||
| 配列の中央値を根とすれば良い。 | ||
| 時間計算量: O(N) => 各要素が1回ずつ処理されるので -> あってる？ |
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リストのスライスを作る計算手間もありますね。https://wiki.python.org/moin/TimeComplexity#:~:text=O(n)-,Get%20Slice,-O(k
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計算量は
( $ 2T(N/2) $ … 半分にした配列を引数にした再帰呼び出し2回分
$ O(N) $ … スライスの分)
を満たすので
arahi10
reviewed
Mar 11, 2026
| ## Step2 | ||
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| - https://github.com/irohafternoon/LeetCode/pull/28/changes#r2061089474 | ||
| >この問題だとHelperを使わない再帰の書き方もありますね。そのほうがコンパクトになり、またtotalに関する議論もなくなるので個人的には好きですが、他の問題ではHelperを使ったほうが見通し良いものもあるので、単なる好みの範疇かもしれません。 |
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そもそもヘルパー関数要らないと思うのですが、ヘルパー関数にした意図って何でしょうか?
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if len(nums) == 1:
return TreeNode(nums[0])
if len(nums) == 0:
return None
mid = len(nums) // 2
root = TreeNode(nums[mid])
left = nums[:mid]
right = nums[mid+1:]
root.left = self.sortedArrayToBST(left)
root.right = self.sortedArrayToBST(right)
return root
arahi10
reviewed
Mar 11, 2026
|
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| 再帰的にサブツリーを構成していけば良いと考えた。サブツリーの具体的な構成方法について | ||
| 配列の中央値を根とすれば良い。 | ||
| 時間計算量: O(N) => 各要素が1回ずつ処理されるので -> あってる？ |
There was a problem hiding this comment.
計算量は
( $ 2T(N/2) $ … 半分にした配列を引数にした再帰呼び出し2回分
$ O(N) $ … スライスの分)
を満たすので
| if not nums: | ||
| return None | ||
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| dummy = TreeNode() |
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この番兵置くなら上の if not num はなくても問題なく動きそうですね。
nodchip
reviewed
Mar 14, 2026
| ```py | ||
| class Solution: | ||
| def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: | ||
| def construct_bintree(left_idx, right_idx): |
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こちらのコメントをご参照ください。
hemispherium/LeetCode_Arai60#10 (comment)
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解く問題
Convert Sorted Array To Binary Search Tree
次に解く問題
Path Sum