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fhiyo
reviewed
Jun 28, 2024
| def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]: | ||
| if not root1: | ||
| return root2 | ||
| elif not root2: |
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そうですね。elif は、結構強い意図を感じます。
if A:
B
elif C:
D
Eとあったときに、B の継続が E であることを強く主張しているように見えます。
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elifってそんな印象になってしまうんですね, ありがとうございます
sakupan102
reviewed
Jun 29, 2024
| # root1に統合 | ||
| class Solution: | ||
| def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]: | ||
| if not root1: |
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全体的に良いと思いました。
どちらかがNoneの場合root1 or root2でreturnしても良いかもしれません。
Mike0121
reviewed
Jun 29, 2024
| root1.val += root2.val | ||
| root1.left = self.mergeTrees(root1.left, root2.left) | ||
| root1.right = self.mergeTrees(root1.right, root2.right) | ||
| return root1 |
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全体的に良いと思いました。dequeを使ってBFSでもかけるので、書いてみても良いと思います。
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問題
617. Merge Two Binary Trees