diff --git a/binary-tree-level-order-traversal/JustHm.swift b/binary-tree-level-order-traversal/JustHm.swift
new file mode 100644
index 000000000..27e99ffdd
--- /dev/null
+++ b/binary-tree-level-order-traversal/JustHm.swift
@@ -0,0 +1,17 @@
+// 재귀 함수로 해결
+// 각 노드의 레벨을 기억하고있다가 정답변수에 레벨별로 노드값을 넣어줌
+class Solution {
+    func levelOrder(_ root: TreeNode?) -> [[Int]] {
+        var answer = [[Int]]()
+        searchNodes(0, root, &answer)
+        return answer
+    }
+
+    func searchNodes(_ level: Int, _ node: TreeNode?, _ answer: inout [[Int]]) {
+        guard let node else { return }
+        if level >= answer.count { answer.append([]) }
+        answer[level].append(node.val)
+        searchNodes(level + 1, node.left, &answer)
+        searchNodes(level + 1, node.right, &answer)
+    }
+}
diff --git a/counting-bits/JustHm.swift b/counting-bits/JustHm.swift
new file mode 100644
index 000000000..c8aa39f6c
--- /dev/null
+++ b/counting-bits/JustHm.swift
@@ -0,0 +1,14 @@
+// time: O(n)
+// approach - 2자리 3자리 ... 로 2진법 변환시 맨 앞 1은 고정이다.
+// 그 이후 나머지는 0~N자리일때 총 1의 갯수 를 더해줬다.
+class Solution {
+    func countBits(_ n: Int) -> [Int] {
+        var answer: [Int] = [0, 1]
+
+        while answer.count < n + 1 {
+            answer += answer.map{$0 + 1}
+        }
+
+        return [Int](answer.prefix(n + 1))
+    }
+}