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nodchip
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Apr 7, 2024
| # self.left = left | ||
| # self.right = right | ||
| class Solution: | ||
| def height(self,root:TreeNode) -> bool: |
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同じノードに対し、同じ処理を複数回行っており、コンピューターリソースがもったいないように思います。キャッシュを撮ってはいかがでしょうか？
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コメントありがとうございます
キャッシュ書いてみたことないのですが、トライしてみます！
hayashi-ay
reviewed
Apr 7, 2024
hayashi-ay
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hayashi-ay
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再帰の処理をよくよく追っていくとheightとisBalancedでノードを同じように辿っているので一緒にしてしまうという選択肢もあると思います。
| def isBalanced(self,root:TreeNode) -> bool: | ||
| if not root: | ||
| return True | ||
| return abs(self.height(root.left)-self.height(root.right)) < 2 and self.isBalanced(root.left) and self.isBalanced(root.right) |
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チェックする順番的に以下の方が自然かなという気がしました。
- 左右の子ノードについてバランスしているか
- 自信のノードについてバランスしているか
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好みのかもしれないですが。
if not self.isBalanced(root.left):
return False
if not self.isBalanced(root.right):
return False
return 自分自信のノードがバランスしているか
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ありがとうございます
・再起をひとまとめにする
・バランスしているかの確認の順番、左右を先にしたほうがよさそうですね
(特に再起をひとまとめにすると、そうしたほうが自然)
・一つのreturn文にまとめすぎると読むの大変(構造見えるようにしたほうが読みやすい)のは、たしかにそうですね
直してみます！
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https://leetcode.com/problems/balanced-binary-tree/