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31 changes: 31 additions & 0 deletions binary-tree-level-order-traversal/freemjstudio.py

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🏷️ 알고리즘 패턴 분석

binary-tree-level-order-traversal/freemjstudio.py
from collections import deque

# 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 levelOrder(self, root: TreeNode | None) -> list[list[int]]:
        visits = []

        # bfs
        queue = deque([])
        queue.append(root)

        while queue:
            temp = []
            for _ in range(len(queue)):
                node = queue.popleft()
                if node:
                    temp.append(node.val)

                if node and node.left:
                    queue.append(node.left)
                if node and node.right:
                    queue.append(node.right)
            if temp:
                visits.append(temp)

        return visits
  • 패턴: BFS, Binary Search
  • 설명: 코드가 BFS를 이용해 트리의 레벨 순으로 노드를 방문합니다. 큐를 사용해 각 레벨의 노드를 순회하고, 방문한 값을 모아 결과를 구성합니다. 페이지를 따라가듯 너비 우선으로 진행합니다.

Original file line number Diff line number Diff line change
@@ -0,0 +1,31 @@
from collections import deque

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복잡도 분석을 주석으로 써주시면 좋을 것 같아요


# 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 levelOrder(self, root: TreeNode | None) -> list[list[int]]:
visits = []

# bfs
queue = deque([])
queue.append(root)

while queue:
temp = []

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temp 보다 조금 더 명시적으로 드러나는 변수명을 써주시면 좋을 것 같습니다~

for _ in range(len(queue)):
node = queue.popleft()
if node:
temp.append(node.val)

if node and node.left:
queue.append(node.left)
if node and node.right:
queue.append(node.right)
Comment on lines +21 to +27

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뒤의 두 if 문을 맨 앞 조건 문 안으로 넣어주면 불필요한 if node 체크 로직을 줄일 수 있어 보이네요.

if temp:
visits.append(temp)

return visits
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