Update 0130.被围绕的区域.md, 增加Python3 版本的DFS解法

 增加Python3 版本的DFS解法，在leetcode上测试通过。

Author: TonyHanzhiSU

problems/0130.被围绕的区域.md

@@ -385,6 +385,55 @@ class Solution {
     }
 }
 ```
+### Python3
+
+```Python
+// 深度优先遍历
+class Solution:
+    dir_list = [(0, 1), (0, -1), (1, 0), (-1, 0)]
+    def solve(self, board: List[List[str]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        row_size = len(board)
+        column_size = len(board[0])
+        visited = [[False] * column_size for _ in range(row_size)]
+        # 从边缘开始，将边缘相连的O改成A。然后遍历所有，将A改成O，O改成X
+        # 第一行和最后一行
+        for i in range(column_size):
+            if board[0][i] == "O" and not visited[0][i]:
+                self.dfs(board, 0, i, visited)
+            if board[row_size-1][i] == "O" and not visited[row_size-1][i]:
+                self.dfs(board, row_size-1, i, visited)
+
+        # 第一列和最后一列
+        for i in range(1, row_size-1):
+            if board[i][0] == "O" and not visited[i][0]:
+                self.dfs(board, i, 0, visited)
+            if board[i][column_size-1] == "O" and not visited[i][column_size-1]:
+                self.dfs(board, i, column_size-1, visited)
+        
+        for i in range(row_size):
+            for j in range(column_size):
+                if board[i][j] == "A":
+                    board[i][j] = "O"
+                elif board[i][j] == "O":
+                    board[i][j] = "X"
+
+    
+    def dfs(self, board, x, y, visited):
+        if visited[x][y] or board[x][y] == "X":
+            return
+        visited[x][y] = True
+        board[x][y] = "A"
+        for i in range(4):
+            new_x = x + self.dir_list[i][0]
+            new_y = y + self.dir_list[i][1]
+            if new_x >= len(board) or new_y >= len(board[0]) or new_x < 0 or new_y < 0:
+                continue
+            self.dfs(board, new_x, new_y, visited)
+
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">