108. Convert Sorted Array to Binary Search Tree #24
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tokuhirat
reviewed
Aug 10, 2025
| root = TreeNode(nums[middle_index]) | ||
| root.left = self.sortedArrayToBST(nums[:middle_index]) | ||
| root.right = self.sortedArrayToBST(nums[middle_index+1:]) | ||
| return root |
huyfififi
approved these changes
Aug 10, 2025
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| ## 他の人のPRやコメントを踏まえて実装 | ||
| leetcodeの解法 | ||
| height-balancedだから空間計算量はO(logn)か |
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インデックスを用いる解法なら、PythonのList Slicingによるコピーが発生しないので、空間計算量はcall stack分のO(logn)になりそうですよね。
ryosuketc
reviewed
Aug 14, 2025
| return result | ||
| ``` | ||
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| indexを狭めていく方式もある |
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step3 の nums 自体をスライスして渡す方と比べると、個人的には index を使ったこちらの方が好みですね。問題の条件によって選択できると良さそうです。
| middle_index = len(nums) // 2 | ||
| root = TreeNode(nums[middle_index]) | ||
| root.left = self.sortedArrayToBST(nums[:middle_index]) | ||
| root.right = self.sortedArrayToBST(nums[middle_index+1:]) |
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これイマイチ明記がなくいつも迷うのですが、インデックスは一応スペースを開けることが多いのではないかと思います (nums[middle_index + 1])。
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問題：108. Convert Sorted Array to Binary Search Tree
次の問題：112. Path Sum