Tic-Tac-Toe code using Python with Tkinter

Let us begin with some layout. This is a typical grid use case so that you won't have to manually manage grid. At the lowest level, a grid structure essentially places buttons by calculating coordinates. But this gets tedious over time, that's why a grid layout is provided. Also, it was a clever way of cheking win by being symbol agnostic. A normal tic tac toe game would have used checkwin(symbol) to determine win.

On strings

To escape ' within ' ' you can use " instead. From

'\'S TURN.'

to

"'S TURN."

and you can also use string formatting to clear up some clutter.

"PLAYER {}'S TURN.".format(turn)

On layout

Modifying your create_button function to this allows a grid structure

def create_button(x, y):
    button = tk.Button(width=10, height=2, command=lambda: callback(button))
    button.grid(row=x, column=y)
    return button

Then we modify others since different layouts can't be mixed

label_button = ttk.Button(
                        game, 
                        text="PLAYER {}'S TURN.".format(turn), 
                        style='Fun.TButton', width=20, 
                        state='disabled')
label_button.grid(row=0, column=1)
invalid = tk.Label(text='')
invalid.grid(row=4, column=1)

adding the buttons can be then done as

buttons = []

buttons.append(create_button(1, 0))
buttons.append(create_button(1, 1))
buttons.append(create_button(1, 2))
buttons.append(create_button(2, 0))
buttons.append(create_button(2, 1))
buttons.append(create_button(2, 2))
buttons.append(create_button(3, 0))
buttons.append(create_button(3, 1))
buttons.append(create_button(3, 2))

You can use a loop for the row and itertools.cycle for the 0, 1, 2 if you want to simplify it.

The check_win function

• Simplifying if

Adding a () to if statements allows you to write or without \

if ... :
    ...

to

if (...):
    ...

thus the win_function can be simplified from

if (buttons[0]['text'] == buttons[1]['text'] == buttons[2]['text'] != '') or \
(buttons[3]['text'] == buttons[4]['text'] == buttons[5]['text'] != '') or \

to

def check_win():
    if (
        (buttons[0]['text'] == buttons[1]['text'] == buttons[2]['text'] != '') or
        (buttons[3]['text'] == buttons[4]['text'] == buttons[5]['text'] != '') or
        ...
        ):
        return True
    else:
        return False

• Simplifying values write-up

This can also be further simplified by defining a function to replace buttons[0]['text']

def btext(i):
    return buttons[i]['text']

and using it

def check_win():
    if (
        (btext(0) == btext(1) == btext(2) != '') or
        (btext(3) == btext(4) == btext(5) != '') or
        (btext(6) == btext(7) == btext(8) != '') or
        (btext(0) == btext(3) == btext(6) != '') or
        (btext(1) == btext(4) == btext(7) != '') or
        (btext(2) == btext(5) == btext(8) != '') or
        (btext(2) == btext(4) == btext(6) != '') or
        (btext(0) == btext(6) == btext(8) != '')
        ):
        return True
    else:
        return False

On architecture

A common pattern is the MVC (Model, View, Controller). While checking and updating gui directly works here, you might consider adding states in a structure like this:

board = [
    ['', '', ''],
    ['', '', ''],
    ['', '', '']
]

Operations are done on this and the gui is updated according to this.

From : Python Tic Tac Toe Game

    win_commbinations = ((0, 1, 2), (3, 4, 5), (6, 7, 8), (0, 3, 6), (1, 4, 7), (2, 5, 8), (0, 4, 8), (2, 4, 6))

for a in win_commbinations:
            if board[a[0]] == board[a[1]] == board[a[2]] == "X":
                print("Player 1 Wins!\n")
                print("Congratulations!\n")
                return True

            if board[a[0]] == board[a[1]] == board[a[2]] == "O":
                print("Player 2 Wins!\n")
                print("Congratulations!\n")
                return True
        for a in range(9):
            if board[a] == "X" or board[a] == "O":
                count += 1
            if count == 9:
                print("The game ends in a Tie\n")
                return True

This alternative solution is a bit cleaner + includes a "Tie" check, which your original solution doesn't check ( just remove if you consider it irrelevant ofc )