如何将寻找矩阵中所有可能路径的递归函数转换为迭代函数？

def find_all_paths_recursive(matrix, x, y, path, all_paths):
    if x == len(matrix) - 1 and y == len(matrix[0]) - 1:
        all_paths.append(path + [(x, y)])
        return
    
    # Mark the current cell as visited
    temp = matrix[x][y]
    matrix[x][y] = -1
    
    # Explore all possible directions: right, down, left, up
    for dx, dy in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
        nx, ny = x + dx, y + dy
        if 0 <= nx < len(matrix) and 0 <= ny < len(matrix[0]) and matrix[nx][ny] != -1:
            find_all_paths_recursive(matrix, nx, ny, path + [(x, y)], all_paths)
    
    # Unmark the current cell
    matrix[x][y] = temp

def find_all_paths(matrix):
    all_paths = []
    find_all_paths_recursive(matrix, 0, 0, [], all_paths)
    return all_paths

def find_all_paths_iterative(matrix):
    if not matrix or not matrix[0]:
        return []
    
    all_paths = []
    stack = [([(0, 0)], 0, 0)]
    
    while stack:
        path, x, y = stack.pop()
        
        if x == len(matrix) - 1 and y == len(matrix[0]) - 1:
            all_paths.append(path + [(x, y)])
            continue
        
        # Mark the current cell as visited
        temp = matrix[x][y]
        matrix[x][y] = -1
        
        # Explore all possible directions: right, down, left, up
        for dx, dy in [(0, 1), (1, 0), (0, -1), (-1, 0)]:
            nx, ny = x + dx, y + dy
            if 0 <= nx < len(matrix) and 0 <= ny < len(matrix[0]) and matrix[nx][ny] != -1:
                stack.append((path + [(x, y)], nx, ny))
        
        # Unmark the current cell
        matrix[x][y] = temp
    
    return all_paths