为什么我的递归python函数返回“ None”？

Question

我正在尝试编写python程序来解决数独难题，并​​且正在使用回溯方法。我已经设法解决了难题，但是当我尝试返回已解决的数独时，它只返回了一个，但是可以打印数独。这是代码的主体，我得到的结果如下：

def sudoku_solver(sudoku_to_solve):

    #solve using backtracking

    sudoku = sudoku_to_solve
    i = 0
    j = 0

    possiblities = {}

    #check if sudoku is full
    if isFull(sudoku) == True:
        print("solved")
        print(sudoku)
        return sudoku
    else:
        #find first 0
        for x in range (0, 9):
            for y in range (0, 9):
                if sudoku[x][y] == 0:
                    i = x
                    j = y
                    break
            else:
                continue
            break

        possiblities = possibleEntries(sudoku, i, j)

        for x in range (1, 10):
            if not possiblities[x] == 0:
                sudoku[i][j] = possiblities[x]
                sudoku_solver(sudoku)
        # backtrack
        sudoku[i][j] = 0

        #return solved_sudoku




print(sudokus[0])
print(sudoku_solver(sudokus[0]))

[[0. 0. 4. 0. 8. 3. 0. 0. 2.]
 [0. 5. 1. 0. 0. 4. 3. 0. 0.]
 [0. 0. 0. 0. 9. 6. 7. 1. 0.]
 [1. 2. 0. 8. 0. 0. 0. 0. 6.]
 [0. 4. 0. 0. 0. 0. 5. 0. 0.]
 [8. 3. 0. 6. 0. 7. 9. 0. 0.]
 [0. 6. 0. 3. 0. 9. 0. 4. 0.]
 [0. 0. 7. 0. 0. 0. 2. 0. 5.]
 [0. 9. 0. 0. 5. 0. 8. 0. 3.]]
solved
[[9. 7. 4. 1. 8. 3. 6. 5. 2.]
 [6. 5. 1. 2. 7. 4. 3. 8. 9.]
 [2. 8. 3. 5. 9. 6. 7. 1. 4.]
 [1. 2. 9. 8. 3. 5. 4. 7. 6.]
 [7. 4. 6. 9. 1. 2. 5. 3. 8.]
 [8. 3. 5. 6. 4. 7. 9. 2. 1.]
 [5. 6. 8. 3. 2. 9. 1. 4. 7.]
 [3. 1. 7. 4. 6. 8. 2. 9. 5.]
 [4. 9. 2. 7. 5. 1. 8. 6. 3.]]
None

任何帮助将不胜感激