diff --git a/non-overlapping-intervals/seungriyou.py b/non-overlapping-intervals/seungriyou.py
new file mode 100644
index 000000000..20ad7aff2
--- /dev/null
+++ b/non-overlapping-intervals/seungriyou.py
@@ -0,0 +1,57 @@
+# https://leetcode.com/problems/non-overlapping-intervals/
+
+from typing import List
+
+class Solution:
+    def eraseOverlapIntervals1(self, intervals: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(nlogn) (sort)
+            - SC: O(1) (tim sort -> 최악의 경우 O(n)까지 가능)
+
+        [Approach]
+            intervals를 start 기준으로 오름차순 정렬 후, greedy 하게 항상 작은 end를 가지는 interval을 선택한다.
+        """
+        # sort asc by start
+        intervals.sort(key=lambda x: x[0])
+
+        # initialize prev_e
+        prev_e = intervals[0][0] - 1
+        to_be_removed = 0
+
+        for s, e in intervals:
+            # (1) overlapping       : greedy하게 prev_e와 e 중 더 작은 값으로 prev_e를 업데이트하고, 제거할 interval 수 증가
+            if prev_e > s:
+                prev_e = min(prev_e, e)
+                to_be_removed += 1
+            # (2) non-overlapping   : prev_e만 업데이트
+            else:
+                prev_e = e
+
+        return to_be_removed
+
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(nlogn) (sort)
+            - SC: O(1) (tim sort -> 최악의 경우 O(n)까지 가능)
+
+        [Approach]
+            이전 풀이의 (1) overlapping 단계에서 prev_e와 e 중 min을 고르는 로직을 제거하려면, intervals를 end 기준으로 오름차순 정렬하면 된다.
+        """
+        # sort asc by end
+        intervals.sort(key=lambda x: x[1])
+
+        # initialize prev_e
+        prev_e = intervals[0][0] - 1
+        to_be_removed = 0
+
+        for s, e in intervals:
+            # (1) overlapping       : greedy하게 prev_e와 e 중 더 작은 값으로 선택하면 되므로 prev_e를 그대로 두고, 제거할 interval 수 증가
+            if prev_e > s:
+                to_be_removed += 1
+            # (2) non-overlapping   : prev_e만 업데이트
+            else:
+                prev_e = e
+
+        return to_be_removed
diff --git a/number-of-connected-components-in-an-undirected-graph/seungriyou.py b/number-of-connected-components-in-an-undirected-graph/seungriyou.py
new file mode 100644
index 000000000..71b32c323
--- /dev/null
+++ b/number-of-connected-components-in-an-undirected-graph/seungriyou.py
@@ -0,0 +1,108 @@
+# https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/
+
+from typing import List
+
+class Solution:
+    def countComponents_bfs(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v + e) (graph)
+
+        [Approach]
+            BFS로 visited를 기록해가며 connected component를 센다.
+        """
+        from collections import deque
+
+        graph = [[] for _ in range(n)]
+        for a, b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+
+        visited, res = set(), 0
+
+        def bfs(start):
+            q = deque([start])
+            visited.add(start)
+
+            while q:
+                pos = q.popleft()
+                for npos in graph[pos]:
+                    if npos not in visited:
+                        q.append(npos)
+                        visited.add(npos)
+
+            return
+
+        for i in range(n):
+            if i not in visited:
+                bfs(i)
+                res += 1
+
+        return res
+
+    def countComponents_dfs(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v + e) (graph)
+
+        [Approach]
+            DFS로 visited를 기록해가며 connected component를 센다.
+        """
+        graph = [[] for _ in range(n)]
+        for a, b in edges:
+            graph[a].append(b)
+            graph[b].append(a)
+
+        visited, res = set(), 0
+
+        def dfs(pos):
+            # 이전에 visited 포함 여부 확인하므로 base condition 생략 가능
+
+            visited.add(pos)
+
+            # recur
+            for npos in graph[pos]:
+                if npos not in visited:
+                    dfs(npos)
+
+            return
+
+        for i in range(n):
+            if i not in visited:
+                dfs(i)
+                res += 1
+
+        return res
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        """
+        [Complexity]
+            - TC: O(v + e * α(v)) (모든 edge & node 한 번씩 탐색)
+            - SC: O(v) (parent, set(...))
+
+        [Approach]
+            edges를 iterate 하며 union-find 수행 후, parent의 종류의 개수를 세면 된다.
+            parent의 종류의 개수를 셀 때는 다시 find_parent(x)로 찾아야 한다!
+        """
+
+        def find_parent(x):
+            if x != parent[x]:
+                parent[x] = find_parent(parent[x])
+            return parent[x]
+
+        def union_parent(x, y):
+            px, py = find_parent(x), find_parent(y)
+
+            if px < py:
+                parent[py] = px
+            else:
+                parent[px] = py
+
+        parent = [i for i in range(n)]
+
+        for x, y in edges:
+            union_parent(x, y)
+
+        return len(set(find_parent(i) for i in range(n)))
diff --git a/remove-nth-node-from-end-of-list/seungriyou.py b/remove-nth-node-from-end-of-list/seungriyou.py
new file mode 100644
index 000000000..1007d23c8
--- /dev/null
+++ b/remove-nth-node-from-end-of-list/seungriyou.py
@@ -0,0 +1,133 @@
+# https://leetcode.com/problems/remove-nth-node-from-end-of-list/
+
+from typing import Optional
+
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def removeNthFromEnd_recur(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        """
+        [Complexity]
+            - TC: O(len)
+            - SC: O(len) (call stack)
+
+        [Approach]
+            재귀적으로 linked list를 확인하며 끝에서부터의 순서를 확인 후, 다음 node의 순서가 n과 같을 때 다음 node를 건너뛰도록 한다.
+            head node를 제거해야 하는 경우(= 끝에서부터 n번째인 node가 head인 경우)에는, 재귀 함수 내에서 node 제거가 불가능하므로 head.next를 반환한다.
+        """
+
+        def check_order_from_end(curr):
+            # base condition
+            if not curr.next:
+                return 1
+
+            # recur
+            next_order_from_end = check_order_from_end(curr.next)
+            # nth from end인 node를 건너뛰기
+            if next_order_from_end == n:
+                curr.next = curr.next.next
+
+            return next_order_from_end + 1
+
+        # head node를 제거해야 하는 경우라면, check_order_from_end() 내에서 node 제거가 불가능하므로 head.next를 반환해야 함
+        if check_order_from_end(head) == n:
+            return head.next
+
+        return head
+
+    def removeNthFromEnd_recur2(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        """
+        [Complexity]
+            - TC: O(len)
+            - SC: O(len) (call stack)
+
+        [Approach]
+            이전 풀이에서 head node를 제거해야 하는 경우를 따로 처리하지 않으려면, 주어진 head를 가리키는 dummy node(= prev)를 추가하고
+            dummy node의 next node를 반환하면 된다.
+        """
+
+        def check_order_from_end(curr):
+            # base condition
+            if not curr.next:
+                return 1
+
+            # recur
+            next_order_from_end = check_order_from_end(curr.next)
+            # nth from end인 node를 건너뛰기
+            if next_order_from_end == n:
+                curr.next = curr.next.next
+
+            return next_order_from_end + 1
+
+        prev = ListNode(next=head)
+        check_order_from_end(prev)
+
+        return prev.next
+
+    def removeNthFromEnd_length(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        """
+        [Complexity]
+            - TC: O(n)
+            - SC: O(1)
+
+        [Approach]
+            linked list의 전체 길이를 구하고, head에서부터 (길이 - n - 1) 번 전진하여 node를 건너뛰면 된다.
+            (2 pass)
+        """
+        # linked list의 length 구하기
+        length = 0
+        curr = head
+        while curr:
+            length += 1
+            curr = curr.next
+
+        # length == n라면, head를 제거
+        if length == n:
+            return head.next
+
+        # length - n - 1 번 이동
+        curr = head
+        for _ in range(length - n - 1):
+            curr = curr.next
+
+        # node 제거
+        curr.next = curr.next.next
+
+        return head
+
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        """
+        [Complexity]
+            - TC: O(len)
+            - SC: O(1)
+
+        [Approach]
+            slow, fast의 two pointer를 이용해 반복문으로 풀 수 있다. (1 pass)
+                1. fast를 n 번 전진
+                2. fast가 끝에 도달한 경우, 첫 번째 node를 제거해야하므로 head.next 반환
+                3. 현재 fast의 위치에서 slow와 fast를 함께 전진하면, fast가 끝에 도달할 때 slow는 뒤에서부터 n + 1번째 node임
+                4. 뒤에서부터 n + 1번째인 node가 n - 1번째 node를 가리키도록 함
+        """
+        slow = fast = head
+
+        # 1. fast를 n 번 전진
+        for _ in range(n):
+            fast = fast.next
+
+        # 2. fast가 끝에 도달한 경우, 첫 번째 node를 제거해야하므로 head.next 반환
+        if not fast:
+            return head.next
+
+        # 3. 현재 fast의 위치에서 slow와 fast를 함께 전진하면,
+        #    fast가 끝에 도달할 때 slow는 뒤에서부터 n + 1번째 node임
+        while fast.next:
+            slow, fast = slow.next, fast.next
+
+        # 4. 뒤에서부터 n + 1번째인 node가 n - 1번째 node를 가리키도록 함
+        slow.next = slow.next.next
+
+        return head
diff --git a/same-tree/seungriyou.py b/same-tree/seungriyou.py
new file mode 100644
index 000000000..ff098a8c6
--- /dev/null
+++ b/same-tree/seungriyou.py
@@ -0,0 +1,34 @@
+# https://leetcode.com/problems/same-tree/
+
+from typing import Optional
+
+# 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:
+        """
+        [Complexity]
+            - TC: O(n)
+            - SC: O(height) (call stack)
+
+        [Approach]
+            재귀적으로 두 tree를 타고 내려가며 확인할 수 있다.
+            각 단계에서 두 tree가 다르다고 판단할 수 있는 조건은
+                (1) 한 쪽 node만 None이거나
+                (2) 두 node의 값이 다른
+            경우이다.
+        """
+        # base condition
+        if not p and not q:
+            return True
+
+        # not same
+        if not p or not q or p.val != q.val:
+            return False
+
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
diff --git a/serialize-and-deserialize-binary-tree/seungriyou.py b/serialize-and-deserialize-binary-tree/seungriyou.py
new file mode 100644
index 000000000..4ff5c51d0
--- /dev/null
+++ b/serialize-and-deserialize-binary-tree/seungriyou.py
@@ -0,0 +1,64 @@
+# https://leetcode.com/problems/serialize-and-deserialize-binary-tree/
+
+# Definition for a binary tree node.
+class TreeNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.left = None
+        self.right = None
+
+class Codec:
+
+    def serialize(self, root):
+        """Encodes a tree to a single string.
+
+        :type root: TreeNode
+        :rtype: str
+        """
+        res = []
+
+        def preorder_ser(node):
+            # base condition
+            if not node:
+                res.append("None")
+                return
+
+            # traversal
+            res.append(str(node.val))
+            preorder_ser(node.left)
+            preorder_ser(node.right)
+
+        preorder_ser(root)
+
+        return ",".join(res)
+
+    def deserialize(self, data):
+        """Decodes your encoded data to tree.
+
+        :type data: str
+        :rtype: TreeNode
+        """
+        from collections import deque
+
+        preorder_vals = deque(data.split(","))
+
+        def preorder_deser():
+            val = preorder_vals.popleft()
+
+            # base condition
+            if val == "None":
+                return None
+
+            # traversal
+            node = TreeNode(int(val))
+            node.left = preorder_deser()
+            node.right = preorder_deser()
+
+            return node
+
+        return preorder_deser()
+
+# Your Codec object will be instantiated and called as such:
+# ser = Codec()
+# deser = Codec()
+# ans = deser.deserialize(ser.serialize(root))