[doh6077] WEEK 15 solutions#2338
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TonyKim9401 merged 11 commits intoDaleStudy:mainfrom Feb 20, 2026
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지난주에 참여를 못 해서 이번 주에 누락된 문제들 추가했습니다 :) |
DaleSeo
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Feb 18, 2026
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깔끔한 풀이 잘 보았습니다. 소소한 피드백을 드렸으니 참고하세요.
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| current_node = queue.popleft() | ||
| current_level_nodes.append(current_node.val) |
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| class Solution: | ||
| def rob(self, nums: List[int]) -> int: | ||
| # Top Down DP |
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Top-Down DP로 잘 작성되었지만, 시간이 되신다면 Bottom-Up DP로 메모리 최적화해를 시도해보셔도 좋을 것 같아요.
| # Time Complexity O(N) | ||
| def kthSmallest(self, root: Optional[TreeNode], k: int) -> int: | ||
| ans = [] | ||
| final = [0] |
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이 곳에 Python의 nonlocal 키워드를 활용해보시기를 추천드립니다. 참고: https://youtu.be/LNWDLXbrZCg
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항상 변수 접근 때문에 리스트만 썼는데 이런 방법도 있군요. 파이썬을 계속 써왔는데 nonlocal 키워드는 써볼 기회가 없었네요.
피드백 달아주셔서 감사합니다!!
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| while q: | ||
| curr = q.popleft() | ||
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| # Only attempt full compare when root values match | ||
| if curr.val == subRoot.val and helper(curr, subRoot): | ||
| return True | ||
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| if curr.left: | ||
| q.append(curr.left) | ||
| if curr.right: | ||
| q.append(curr.right) |
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이 문제는 보통 재귀만으로 해결하는 경우가 많은데, 이렇게 BFS로 각 노드를 순회하며 서브트리 비교를 하는 방법이 신선하네요.
Co-authored-by: Dale Seo <5466341+DaleSeo@users.noreply.github.com>
Co-authored-by: Dale Seo <5466341+DaleSeo@users.noreply.github.com>
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