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shining-ai
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Feb 5, 2024
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| if not root1 and not root2: | ||
| return None |
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この条件は不要ではないでしょうか。
両方Noneのときは、
if not root1:
return root2
で処理できると思います。
Comment on lines
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| new_node = TreeNode(root1.val + root2.val) | ||
| new_node.left = self.mergeTrees(root1.left, root2.left) | ||
| new_node.right = self.mergeTrees(root1.right, root2.right) |
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新しい木を作るのではなく、既存の木のどちらかに足し算して編集していくほうが自然な流れだと思いました。
好みの問題かもしれませんが。
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問題文的には既存の木に足していく方が自然かもですね。
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https://leetcode.com/problems/merge-two-binary-trees/description/