diff --git a/remove-nth-node-from-end-of-list/mike2ox.ts b/remove-nth-node-from-end-of-list/mike2ox.ts
new file mode 100644
index 000000000..74e9348be
--- /dev/null
+++ b/remove-nth-node-from-end-of-list/mike2ox.ts
@@ -0,0 +1,31 @@
+/**
+ * Source: https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+ * Solution: 두 개의 포인터를 이용해 n번째 노드를 찾아 삭제
+ *
+ * 시간복잡도: O(N) - 두 개의 포인터가 한번씩 순회
+ * 공간복잡도: O(1) - 상수만큼의 공간 사용
+ */
+
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+  if (!head) return null;
+
+  const dummy = new ListNode(0, head);
+  let slow: ListNode | null = dummy;
+  let fast: ListNode | null = dummy;
+
+  for (let i = 0; i <= n; i++) {
+    if (!fast) return head;
+    fast = fast.next;
+  }
+
+  while (fast) {
+    slow = slow!.next;
+    fast = fast.next;
+  }
+
+  if (slow && slow.next) {
+    slow.next = slow.next.next;
+  }
+
+  return dummy.next;
+}
diff --git a/same-tree/mike2ox.ts b/same-tree/mike2ox.ts
new file mode 100644
index 000000000..8ad9cbe45
--- /dev/null
+++ b/same-tree/mike2ox.ts
@@ -0,0 +1,60 @@
+/**
+ * Source: https://leetcode.com/problems/same-tree/
+ * Solution: 트리의 노드를 순회하면서 값이 같은지 확인
+ * 시간 복잡도: O(N) - 트리의 모든 노드를 한번씩 방문
+ * 공간 복잡도: O(N) - 스택에 최대 트리의 높이만큼 쌓일 수 있음
+ *
+ * 추가 사항
+ * - 트리를 순회만 하면 되기에 Typescript로 Stack을 활용해 DFS로 해결
+ * - 재귀로 구현하면 간단하게 구현 가능
+ */
+
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+  if (!q || !p) return !q === !p;
+  let result = true;
+  let stack = new Array({
+    left: p,
+    right: q,
+  });
+  while (stack.length) {
+    const now = stack.pop();
+    const left = now?.left;
+    const right = now?.right;
+    const isLeafNode =
+      !left?.left && !left?.right && !right?.right && !right?.left;
+    const isSameValue = left?.val === right?.val;
+    const hasDifferentSubtree =
+      (!left?.left && right?.left) || (!left?.right && right?.right);
+    if (isLeafNode && isSameValue) continue;
+    if (!isSameValue || hasDifferentSubtree) {
+      result = false;
+      break;
+    }
+    stack.push({ left: left?.left, right: right?.left });
+    stack.push({ left: left?.right, right: right?.right });
+  }
+  return result;
+}
+
+// Solution 2 - 재귀
+function isSameTree2(p: TreeNode | null, q: TreeNode | null): boolean {
+  if (!p && !q) return true;
+  if (!p || !q) return false;
+  if (p.val !== q.val) return false;
+
+  return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+}