From ab6622cc01396a21fcde2e5a1cdcc3a27cdd5543 Mon Sep 17 00:00:00 2001
From: jamiebase <100mgml@gmail.com>
Date: Sat, 23 May 2026 17:41:32 +0900
Subject: [PATCH 1/2] Add solution for checking if two binary trees are the
 same

---
 same-tree/jamiebase.py | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 same-tree/jamiebase.py

diff --git a/same-tree/jamiebase.py b/same-tree/jamiebase.py
new file mode 100644
index 0000000000..08b0bd96e1
--- /dev/null
+++ b/same-tree/jamiebase.py
@@ -0,0 +1,26 @@
+"""
+# Approach
+재귀를 이용해서 왼쪽, 오른쪽 서브트리의 값을 비교합니다
+
+# Complexity
+- Time complexity: 노드 개수 N일 때, O(N)
+- Space complexity: 트리 높이 H 일때, O(H)
+"""
+
+
+# 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 not p and not q:
+            return True
+        if not p or not q:
+            return False
+        if p.val != q.val:
+            return False
+
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

From 555369f5ac0d430f694fdbe8d67306adbea3f7ef Mon Sep 17 00:00:00 2001
From: jamiebase <100mgml@gmail.com>
Date: Tue, 26 May 2026 12:04:35 +0900
Subject: [PATCH 2/2] add: implement lowest common ancestor for binary search
 tree

---
 .../jamiebase.py                              | 22 +++++++++++++++++++
 1 file changed, 22 insertions(+)
 create mode 100644 lowest-common-ancestor-of-a-binary-search-tree/jamiebase.py

diff --git a/lowest-common-ancestor-of-a-binary-search-tree/jamiebase.py b/lowest-common-ancestor-of-a-binary-search-tree/jamiebase.py
new file mode 100644
index 0000000000..a91622b097
--- /dev/null
+++ b/lowest-common-ancestor-of-a-binary-search-tree/jamiebase.py
@@ -0,0 +1,22 @@
+"""
+# Approach
+BST 의 특징을 이용해서 값의 대소 비교로 공통 조상을 찾습니다.
+
+# Complexity
+- Time complexity: 트리의 높이 H일때, O(H)
+- Space complexity: O(1)
+"""
+
+
+class Solution:
+    def lowestCommonAncestor(
+        self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
+    ) -> "TreeNode":
+        cur = root
+        while cur:
+            if p.val < cur.val and q.val < cur.val:
+                cur = cur.left
+            elif p.val > cur.val and q.val > cur.val:
+                cur = cur.right
+            else:
+                return cur