Solved Arai60/108. Convert Sorted Array to Binary Search Tree #24
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oda
reviewed
Jun 5, 2025
| if middle + 1 < right: | ||
| node.right = sortedArrayToBSTHelper(TreeNode(), middle + 1, right) | ||
| return node | ||
| return sortedArrayToBSTHelper(TreeNode(), 0, len(nums)) |
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本当ですね。後からreturnしてくれば先につなげなくてもいいですね
fhiyo
reviewed
Jun 6, 2025
| while node_and_range: | ||
| node, left_index, right_index = node_and_range.popleft() | ||
| middle_index = (left_index + right_index) // 2 | ||
| node.val = nums[middle_index] |
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一応、numsが空配列の場合ここでIndexErrorになりますかね。別解ではNoneを返しているので意図通りかは少し気になりました
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空配列は問題文の条件にはないのですが、他の人のを見ると気にしていない人が多かったので書かないことにしました。
quinn-sasha
reviewed
Jun 24, 2025
| root: TreeNode = TreeNode(val=nums[0]) | ||
| nodes_to_add: deque[TreeNode] = deque([root]) | ||
| for i in range(1, len(nums)): | ||
| node = nodes_to_add.popleft() |
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nodes_to_add という変数名で、同じノードを入れたり出したりしているのがわかりづらいと思いました。
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nodes_to_processとかの方が良いですかね
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問題文：https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/