You are making a common mistake, and looking at the non winning doors as individual doors, don't do that. The Non Winners, or goats, must be treated as the same, not door B, or C or whatever, or as two separate prizes, but as losing choices, period. That way, when you construct your tree, you don't count the same scenario twice, which would skew your answer.

For instance, if door A was the winner, and you picked it, Monty would open either door B or C. If he opened B, and you switched to C, you lose. If he opened C, and you switched to B, you lose. THESE MUST NOT BE COUNTED AS TWO DiFFERENT POSSIBILITIES!!!! It should be one possibility, you picked, you were shown a loser, and you switched to a loser. That will straighten out your tree.

(your way of thinking may be correct if there were three different prizes and you were asked whatr is the probability of getting each prize.)