diff --git a/non-overlapping-intervals/Seoya0512.py b/non-overlapping-intervals/Seoya0512.py
new file mode 100644
index 0000000000..865fa28051
--- /dev/null
+++ b/non-overlapping-intervals/Seoya0512.py
@@ -0,0 +1,23 @@
+'''
+Time Complexity: O(n log n)
+- intervals 배열을 정렬하는데 소요되는 시간 (n은 intervals의 길이)
+- 배열을 순회하며 겹치는 구간을 찾는데 소요되는 시간 O(n)이나 전체 시간 복잡도에 영향을 주지 않음
+
+# Space Complexity: O(1)
+- last_end, count 변수를 사용하는데 필요한 공간
+'''
+
+
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        intervals.sort(key=lambda x: x[1])
+        last_end = intervals[0][1]
+        count = 0
+
+        for val in intervals[1:]:
+            if (val[0] < last_end):
+                print(val)
+                count += 1
+            if (val[0] >= last_end):
+                last_end = val[1]
+        return count
diff --git a/number-of-connected-components-in-an-undirected-graph/Seoya0512.py b/number-of-connected-components-in-an-undirected-graph/Seoya0512.py
new file mode 100644
index 0000000000..9f0ab6cd11
--- /dev/null
+++ b/number-of-connected-components-in-an-undirected-graph/Seoya0512.py
@@ -0,0 +1,43 @@
+'''
+Time Complexity: O(N + E)
+- 노드의 개수 N, 간선의 개수 E
+- 노드의 개수와 간선의 수를 합친만큼 재귀 함수 호출 필요 
+
+Space Complexity: O(N + E)
+- 인접 리스트의 크기가 노드와 간선의 수의 합에 비례 
+- 집합에 최대 N개의 숫자 저장
+'''
+from typing import (
+    List,
+)
+
+class Solution:
+    """
+    @param n: the number of vertices
+    @param edges: the edges of undirected graph
+    @return: the number of connected components
+    """
+    def count_components(self, n: int, edges: List[List[int]]) -> int:
+        # 연결관계를 표현하기 위한 그래프
+        graph = [[] for _ in range(n)]
+
+        for a,b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+        
+        # 깊은탐색에서 노드 방문여부 표기 
+        visited = set()
+
+        def dfs(node):
+        # 재귀함수 실행시 방문 처리
+            visited.add(node)
+            for adj in graph[node]:
+                if adj not in visited:
+                    dfs(adj)
+        
+        count = 0
+        for node in range(n):
+            if node not in visited:
+                dfs(node)
+                count += 1
+        return count
diff --git a/remove-nth-node-from-end-of-list/Seoya0512.py b/remove-nth-node-from-end-of-list/Seoya0512.py
new file mode 100644
index 0000000000..b472927267
--- /dev/null
+++ b/remove-nth-node-from-end-of-list/Seoya0512.py
@@ -0,0 +1,42 @@
+class Solution:
+    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
+        
+        def remove_node(node):
+            if node is None:
+                return 0
+            # 2. 재귀 호출
+            count = remove_node(node.next) + 1
+            # 4. 삭제 로직 (count가 n+1일 때)
+            if count == n + 1:
+                node.next = node.next.next
+            # 5. 현재 count 반환
+            return count
+
+        # 재귀 함수 호출 시작
+        total_count = remove_node(head)
+        
+        if total_count == n:
+            return head.next
+        return head
+
+
+# Two Pointer Solution
+# One loop 
+
+class Solution:
+    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
+        first = head
+        # 삭제할 노드의 바로 전 노드에 도착
+        for _ in range(n+1):
+            first = first.next
+        
+        dummy = ListNode(None, head)
+        second = dummy 
+
+        while first:
+            first = first.next
+            second = second.next
+
+        second.next = second.next.next
+
+        return dummy.next
diff --git a/same-tree/Seoya0512.py b/same-tree/Seoya0512.py
new file mode 100644
index 0000000000..ed74763a82
--- /dev/null
+++ b/same-tree/Seoya0512.py
@@ -0,0 +1,18 @@
+# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if p == None and q == None:
+            return True
+
+        if p == None or q == None:
+            return False
+        
+        if p.val != q.val:
+            return False
+        
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right,q.right)