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[reeseo3o] WEEK 15 Solutions #2638

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week15: subtree-of-another-tree
reeseo3o Jun 12, 2026
52aa51d
style: fix lint in subtree-of-another-tree
reeseo3o Jun 12, 2026
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18 changes: 18 additions & 0 deletions subtree-of-another-tree/reeseo3o.js
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@dalestudy dalestudy Bot Jun 12, 2026

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🏷️ 알고리즘 패턴 분석

  • 패턴: DFS
  • 설명: 이 코드는 재귀 호출을 통해 트리의 모든 노드를 방문하며 서브트리 여부를 검사하는 방식으로, 깊이 우선 탐색(DFS) 패턴에 속합니다.

📊 시간/공간 복잡도 분석

복잡도
Time O(n * m)
Space O(h + m)

피드백: 전체 트리의 각 노드에 대해 서브트리 일치 여부를 검사하는데, 각 검사에서 두 트리의 노드 수에 비례하는 시간이 소요됩니다. 따라서 최악의 경우 시간 복잡도는 O(n * m)입니다.

개선 제안: 현재 구현이 적절해 보입니다.

💡 풀이에 시간/공간 복잡도를 주석으로 남겨보세요!

Original file line number Diff line number Diff line change
@@ -0,0 +1,18 @@
// Time Complexity: O(n*m)
// Space Complexity: O(n+m)

const isSubtree = (root, subRoot) => {
const isSame = (a, b) => {
if (!a && !b) return true;
if (!a || !b) return false;
if (a.val !== b.val) return false;
return isSame(a.left, b.left) && isSame(a.right, b.right);
};

if (!root) return false;
return (
isSame(root, subRoot) ||
isSubtree(root.left, subRoot) ||
isSubtree(root.right, subRoot)
);
};
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