CREATE TABLE Apartment_Buildings (building_address VARCHAR(255), building_id VARCHAR(255)); CREATE TABLE Apartments (building_id VARCHAR(255), bathroom_count INTEGER)
Show the addresses of the buildings that have apartments with more than 2 bathrooms.
    
SQL: SELECT T1.building_address FROM Apartment_Buildings AS T1 JOIN Apartments AS T2 ON T1.building_id = T2.building_id WHERE T2.bathroom_count > 2