It does not matter to regard them as 1 door with price and 2 without. I added the two doors in the tree, to show all options, as any good probability tree should.

Another approach that leads often to the same conclusion is to think about the following:
The probability that my initial chosen door is correct = 1/3
The probability that my initial chosen door is incorrect = 2/3

If my initial chosen door is correct and I switch I will loose, if my initial chosen door is incorrect and I switch I will win. Hence probability of winning with switching = 2/3, loosing 1/3.

Not switching will actually 'switch' these probabilities around.

As mentioned earlier, the full rules are often left out in order to keep things clear. 'The rules' I mean as in that the host knows behind what door the price is, that he will always first open a door that does neither contain the price, neither the original chosen door etc.

If all of this still doesn't convince you, either create a software programe yourself and run it a few times (you could actually do this in Excel even). :-)