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Vivelapaix
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Jun 2, 2020
| :rtype: List[int] | ||
| """ | ||
| head = root | ||
| inorderlist = [] |
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Очень круто, но много кода. Ты такой код не успеешь написать быстро, много разных if. Можешь оставить
Но напиши рядом рекурсивное решение и решение, где ты рекурсию раскроешь через stack, т.е. получится итеративное решение без лишних if
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Наличие многих if - место для потенциальной ошибки. Решения здесь проще
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@Vivelapaix добавил два решения. Ещё вопрос по гиту. Как изменить имя запушенного коммита?
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Jun 4, 2020
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Можно было все в одном блоке python сделать, а не разбивать на 3. Комментариями бы решения отделил.
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Added LEETCODE "binary-tree-inorder-traversal" iterative, nondestructive, stack-free, tag-free left inorder solution.