From beccb8466941911402436e0c6fefb885f36e0cd5 Mon Sep 17 00:00:00 2001
From: Yunyoung Chung <double.y.0525@gmail.com>
Date: Sat, 6 Jun 2026 00:00:34 +0900
Subject: [PATCH 1/2] [7th batch] week - binary tree level order traversal

---
 binary-tree-level-order-traversal/liza0525.py | 56 +++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 binary-tree-level-order-traversal/liza0525.py

diff --git a/binary-tree-level-order-traversal/liza0525.py b/binary-tree-level-order-traversal/liza0525.py
new file mode 100644
index 0000000000..350a6c6e46
--- /dev/null
+++ b/binary-tree-level-order-traversal/liza0525.py
@@ -0,0 +1,56 @@
+# 7기 풀이
+class Solution:
+    # 풀이 1 - BFS
+    # 시간 복잡도: O(n)
+    # - 노드의 개수(n)만큼 모두 탐색하므로 
+    # 공간 복잡도: O(w)
+    # - 최악은 한 레벨의 최대 노드 수(w)만큼 queue에 쌓임
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        res = []
+
+        nodes = deque()
+        if root:
+            nodes.appendleft(root)
+
+        while nodes:
+            childs = deque()
+            sibling_vals = []
+            while nodes:
+                node = nodes.pop()
+                if not node:
+                    continue
+
+                sibling_vals.append(node.val)
+                if node.left:
+                    childs.appendleft(node.left)
+                if node.right:
+                    childs.appendleft(node.right)
+
+            res.append(sibling_vals)
+            nodes = childs
+        
+        return res
+
+    # 풀이 2 - DFS
+    # 시간 복잡도: O(n)
+    # - 노드의 개수(n)만큼 모두 탐색하므로 
+    # 공간 복잡도: O(h)
+    # - 재귀 스택이 최대 나무의 높이(h)만큼 쌓임
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        res = []
+
+        def dfs(node, depth):
+            if not node:
+                return
+            
+            if len(res) == depth:
+                res.append([node.val])
+            else:
+                res[depth].append(node.val)
+
+            dfs(node.left, depth + 1)
+            dfs(node.right, depth + 1)
+
+        dfs(root, 0)
+
+        return res

From f4e8e3f2474cadfb2bffd5d1de83e0a17e5da6e7 Mon Sep 17 00:00:00 2001
From: Yunyoung Chung <double.y.0525@gmail.com>
Date: Sat, 6 Jun 2026 00:00:51 +0900
Subject: [PATCH 2/2] [7th batch] week 14 - counting bits

---
 counting-bits/liza0525.py | 16 ++++++++++++++++
 1 file changed, 16 insertions(+)
 create mode 100644 counting-bits/liza0525.py

diff --git a/counting-bits/liza0525.py b/counting-bits/liza0525.py
new file mode 100644
index 0000000000..7e75f43b75
--- /dev/null
+++ b/counting-bits/liza0525.py
@@ -0,0 +1,16 @@
+# 7기 풀이
+# 시간 복잡도: O(n)
+# - n의 크기만큼 모든 수에 대한 dp 값을 찾음
+# 공간 복잡도: O(n)
+# - n의 크기만큼 dp 어레이를 사용
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        dp = [0 for _ in range(n + 1)]
+
+        for i in range(1, n + 1):
+            if i % 2 == 1:
+                dp[i] = dp[i - 1] + 1
+            else:
+                dp[i] = dp[i // 2]
+
+        return dp