diff --git a/binary-tree-level-order-traversal/Chaedie.py b/binary-tree-level-order-traversal/Chaedie.py
new file mode 100644
index 000000000..474cc0e2e
--- /dev/null
+++ b/binary-tree-level-order-traversal/Chaedie.py
@@ -0,0 +1,22 @@
+"""
+Solution: BFS
+Time: O(n)
+Space: O(n)
+"""
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:        
+        if not root: return []
+
+        q = deque([root])
+        result = []
+        while q:
+            level = []
+            for i in range(len(q)):
+                node = q.popleft()
+                level.append(node.val)
+                if node.left: q.append(node.left)
+                if node.right: q.append(node.right)
+            result.append(level)
+
+        return result
diff --git a/counting-bits/Chaedie.py b/counting-bits/Chaedie.py
new file mode 100644
index 000000000..4ecc7b5cf
--- /dev/null
+++ b/counting-bits/Chaedie.py
@@ -0,0 +1,59 @@
+
+"""
+Solution: 
+    1) bin() 메서드로 binary 만들어주고 
+    2) 1 의 갯수를 세어준다.
+Time: O(n^2)
+Space: O(1)
+"""
+
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        result = [0 for i in range(n + 1)]
+
+        for i in range(n+1):
+            b = bin(i)
+            count = 0
+            for char in b:
+                if char == '1':
+                    count += 1
+            result[i] = count
+        
+        return result
+    
+"""
+Solution: 
+    1) 2로 나눈 나머지가 1bit 이라는 성질을 이용해서 count
+Time: O(n logn)
+Space: O(1)
+"""
+
+class Solution:
+    def countBits(self, n: int) -> List[int]:
+        def count(num):
+            count = 0
+            while num > 0:
+                count += num % 2 
+                num = num // 2
+            return count
+        
+        return [count(i) for i in range(n+1)]
+
+"""
+Solution:
+    1) LSB 가 0 1 0 1 반복되므로 num % 2 를 사용한다.
+    2) 나머지 빗은 LSB를 제외한 값이므로 num // 2 를 사용한다.
+Time: O(n)
+Space: O(1)
+"""
+
+class Solution:
+    
+    def countBits(self, n: int) -> List[int]:
+        dp = [0 for i in range(n+1)]
+
+        for i in range(1, n+1):
+            LSB = i % 2
+            dp[i] = dp[i // 2] + LSB
+
+        return dp