在没有递归的情况下找到给定布尔2D矩阵的孤岛问题

Question

我有代码可以找到二进制二维矩阵中的孤岛数。
如果我使用复杂的2d矩阵，则结果为错误“ RecursionError：比较中超出了最大递归深度”。因此，我需要不递归地实施DFS。

# Program to count islands in boolean 2D matrix 
class Graph: 

    def __init__(self, row, col, g): 
        self.ROW = row 
        self.COL = col 
        self.graph = g 

    # A function to check if a given cell  
    # (row, col) can be included in DFS 
    def isSafe(self, i, j, visited): 
        # row number is in range, column number 
        # is in range and value is 1  
        # and not yet visited 
        return (i >= 0 and i < self.ROW and 
                j >= 0 and j < self.COL and 
                not visited[i][j] and self.graph[i][j]) 


    # A utility function to do DFS for a 2D  
    # boolean matrix. It only considers 
    # the 8 neighbours as adjacent vertices 
    def DFS(self, i, j, visited): 

        # These arrays are used to get row and  
        # column numbers of 8 neighbours  
        # of a given cell 
        rowNbr = [-1, -1, -1,  0, 0,  1, 1, 1]; 
            colNbr = [-1,  0,  1, -1, 1, -1, 0, 1]; 

        # Mark this cell as visited 
        visited[i][j] = True

        # Recur for all connected neighbours 
        for k in range(8): 
            if self.isSafe(i + rowNbr[k], j + colNbr[k], visited): 
                self.DFS(i + rowNbr[k], j + colNbr[k], visited) 


    # The main function that returns 
    # count of islands in a given boolean 
    # 2D matrix 
    def countIslands(self): 
        # Make a bool array to mark visited cells. 
        # Initially all cells are unvisited 
        visited = [[False for j in range(self.COL)]for i in range(self.ROW)] 

        # Initialize count as 0 and travese  
        # through the all cells of 
        # given matrix 
        count = 0
        for i in range(self.ROW): 
            for j in range(self.COL): 
                # If a cell with value 1 is not visited yet,  
                # then new island found 
                if visited[i][j] == False and self.graph[i][j] ==1: 
                    # Visit all cells in this island  
                    # and increment island count 
                    self.DFS(i, j, visited) 
                    count += 1

        return count