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[yuseok89] WEEK 10 Solutions #2836

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21 changes: 21 additions & 0 deletions invert-binary-tree/yuseok89.py
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@dalestudy dalestudy Bot Aug 24, 2026

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🏷️ 알고리즘 패턴 분석

invert-binary-tree/yuseok89.py 
# TC: O(N)
# SC: O(N)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:

        if not root:
            return None
        else:
            cur_node = TreeNode(root.val)

            cur_node.right = self.invertTree(root.left)
            cur_node.left = self.invertTree(root.right)

            return cur_node
  • 패턴: Binary Search, Depth-First Search, Divide and Conquer
  • 설명: 이 코드는 이진트리의 자식 노드를 서로 바꿔 트리를 반전시키는 재귀적 분할 처리로서, DFS 방식으로 각 서브트리를 처리하며 분할해서 해결하는 패턴을 사용합니다.

📊 시간/공간 복잡도 분석

유저 분석 실제 분석 결과
Time O(N) O(n) ✅
Space O(N) O(h) ❌

피드백: 트리의 각 노드를 한 번씩 방문하여 자식 정보를 교환하므로 시간 복잡도는 O(n)이고, 재귀 호출 스택의 깊이에 비례하는 공간 복잡도 O(h)이다.

개선 제안: 현재 구현이 적절해 보입니다.

@alphaorderly alphaorderly Aug 24, 2026

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앞의 if가 이미 리턴을 하고 있어서 else 를 크게 쓰지 않으셔도 되지 않을까? 라는 생각이 듭니다!
또한 cur_node 를 만들지 않으시고 root를 직접 바꾸셔도 정답처리 되실거에요!

Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
# TC: O(N)
# SC: O(N)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:

if not root:
return None
else:
cur_node = TreeNode(root.val)

cur_node.right = self.invertTree(root.left)
cur_node.left = self.invertTree(root.right)

return cur_node

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