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[jamiebase] WEEK 13 Solutions #2615
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ab6622c
Add solution for checking if two binary trees are the same
jamiebase 606e0eb
Merge branch 'DaleStudy:main' into main
jamiebase 15012d8
Merge branch 'DaleStudy:main' into main
jamiebase 555369f
add: implement lowest common ancestor for binary search tree
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22 changes: 22 additions & 0 deletions
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lowest-common-ancestor-of-a-binary-search-tree/jamiebase.py
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,22 @@ | ||
| """ | ||
| # Approach | ||
| BST 의 특징을 이용해서 값의 대소 비교로 공통 조상을 찾습니다. | ||
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| # Complexity | ||
| - Time complexity: 트리의 높이 H일때, O(H) | ||
| - Space complexity: O(1) | ||
| """ | ||
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| class Solution: | ||
| def lowestCommonAncestor( | ||
| self, root: "TreeNode", p: "TreeNode", q: "TreeNode" | ||
| ) -> "TreeNode": | ||
| cur = root | ||
| while cur: | ||
| if p.val < cur.val and q.val < cur.val: | ||
| cur = cur.left | ||
| elif p.val > cur.val and q.val > cur.val: | ||
| cur = cur.right | ||
| else: | ||
| return cur |
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🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 트리의 높이 H에 비례하는 시간 복잡도를 가지며, 반복문을 통해 상위 노드로 이동하므로 공간 복잡도는 상수입니다.
개선 제안: 현재 구현이 적절해 보입니다.