Number of Closed Islands LeetCode solution in Python

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class Solution:
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    """主函数：计算封闭岛屿的数量"""
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    def closedIsland(self, grid: List[List[int]]) -> int:
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        result = 0
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        m, n = len(grid), len(grid[0])
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        self.direction = [[1, 0], [-1, 0], [0, 1], [0, -1]]
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        # 遍历 grid，处理边缘陆地
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        for j in range(n):
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            self.dfs(grid, 0, j)
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            self.dfs(grid, m - 1, j)
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        for i in range(m):
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            self.dfs(grid, i, 0)
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            self.dfs(grid, i, n - 1)
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        # 剩下都是封闭岛屿，遍历找结果
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        for i in range(m):
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            for j in range(n):
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                if grid[i][j] == 0:
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                    result += 1
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                    self.dfs(grid, i, j)
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        return result
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    """从 (i, j) 开始，将与之相邻的陆地都变成海水"""
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    def dfs(self, grid, i, j):
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        m, n = len(grid), len(grid[0])
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        # 超出索引边界
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        if i < 0 or j < 0 or i >= m or j >= n:
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            return
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        # 已经是海水了
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        if grid[i][j] == 1:
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            return
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        # 变成海水
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        grid[i][j] = 1
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        for d in self.direction:
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            x = i + d[0]
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            y = j + d[1]
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            self.dfs(grid, x, y)