105. Construct Binary Tree from Preorder and Inorder Traversal #43
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変数の命名も含め読みやすいコードだと思いました。 |
nodchip
reviewed
Mar 8, 2024
| root_index_in_inorder = inorder.index(root_value) | ||
| num_of_left_node = root_index_in_inorder | ||
| root = TreeNode(root_value) | ||
| root.left = self.buildTree(preorder[1:1 + num_of_left_node], inorder[:root_index_in_inorder]) |
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細かい点となりますが、「変数 + 1」と「1 + 変数」は、「変数 + 1」 に統一したほうが読みやすいかなと思いました。
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「インデックスの位置 + ずらす個数」という役割で順番を決めていましたが、指摘された内容も確かにと思いました。
「1 + num_of_left_node」は、rootの位置の隣（1）からnum_of_left_node分
「inorder[root_index_in_inorder + 1:]」はroot_index_in_inorderの位置から1つ分
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| def build_tree_in_preorder(left, right): | ||
| nonlocal root_index | ||
| if left >= right: |
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仰る通りですが、 if left == rightのみだとif left > rightのケースが考慮しなくて良いということを読み手が確認するひと手間が発生するかなと思いif left >= rightにしました。
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https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/