diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/1219. Path with Maximum Gold" "b/leetcode3/\354\265\234\354\233\220\354\244\200/1219. Path with Maximum Gold"
new file mode 100644
index 00000000..0ced48ab
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/1219. Path with Maximum Gold"	
@@ -0,0 +1,59 @@
+#
+
+'''
+1. 아이디어 :
+모든 선택은 상태값을 들고 있어야한다.
+재귀 백트래킹을 통해 롤백.
+
+2. 시간복잡도 :
+    O(4 ** 25)
+
+3. 자료구조/알고리즘 :
+dfs, backtracking
+
+'''
+class Solution:
+    def getMaximumGold(self, grid: list[list[int]]) -> int:
+        n = len(grid)
+        m = len(grid[0])
+        # n <= 15
+        # 비싼쪽으로X 다익스트라X 백트래킹? 4**25 압축?
+
+        dx = [0,0,1,-1]
+        dy = [1,-1,0,0]
+
+        def dfs(row, col):
+            gold = grid[row][col]
+            grid[row][col] = 0 #
+            
+            max_gold = 0
+            for i in range(4):
+                nx = row + dx[i]
+                ny = col + dy[i]
+
+                if 0<= nx <n and 0<=ny < m and grid[nx][ny] != 0:
+                    max_gold = max(max_gold, dfs(nx, ny))
+
+            grid[row][col] = gold
+
+            return gold + max_gold
+
+        ans = 0
+        for i in range(n):
+            for j in range(m):
+                ans = max(ans, dfs(i, j))
+        return ans
+
+'''
+ [1,0,7],
+ [2,0,6],
+ [3,4,5],
+ [0,3,0],
+ [9,0,20]
+
+ [ 6, 0,18]
+ [ 6, 0,18]
+ [ 6,-1,18]
+ [ 0, 3, 0]
+ [ 0, 0, 0]
+ '''
diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/543. Diameter of Binary Tree.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/543. Diameter of Binary Tree.py"
new file mode 100644
index 00000000..4037db92
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/543. Diameter of Binary Tree.py"	
@@ -0,0 +1,35 @@
+# 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
+#
+
+'''
+1. 아이디어 :
+- 현재 노드를 포함한게 가장 길수도 있기에 왼쪽/오른쪽 중 깊은것 리턴
+- 현재 노드 자식들을 계산한게 가장 길수도 있기에 왼쪽 깊이 + 오른쪽 깊이를 리턴
+
+
+2. 시간복잡도 :
+    O(n)
+
+3. 자료구조/알고리즘 :
+dfs
+
+'''
+
+class Solution:
+    def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
+
+        def max_length(node): #deepest, longest
+            if not node:
+                return 0, 0
+            
+            left_deepest, left_longest = max_length(node.left)
+            right_deepest, right_longest = max_length(node.right)
+
+            return max(left_deepest, right_deepest) + 1, max(left_longest, right_longest, left_deepest + right_deepest)
+
+        return max_length(root)[1]