215 Bradford Street, San Francisco, CA 94110
- 3 beds |
- 2 baths |
- 1,334 sqft
Single-Family Home
Built in 1960
1,751sqft lot
2 car garage
$971/sqft
Listed 5 days ago
Property Description
Extensively remodeled Bernal Heights home in a very walkable location, tucked away just off Cortland Ave. 3BR/2BA, approx. 1,334 sq. ft. (per graphic artist), featuring a stunning open-concept renovation. The main level showcases a light-filled kitchen with an island offering dual prep space, a custom pantry with pull-out drawers, a Bosch appliance suite, and a skylight illuminating the kitchen and dining areas. Custom built-in seating was designed to fit the dining space beautifully. The living room features a gas fireplace and windows framing serene sunset views of Bernal Hill. Hardwood floors run throughout, and both upper bedrooms enjoy Bay and Oakland skyline views. Downstairs, a third bedroom and full bathroom create excellent separation, ideal as a primary suite or guest retreat, with sliding doors opening to the sun-drenched backyard. A 2-car tandem garage offers interior and backyard access. Walk Score of 91. Cortland Ave is a 9-minute stroll, Barebottle Brewing just 1 minute away. Close to Black Jet Baking Co., Martha & Bros. Coffee, United Dumplings, and Good Life Grocery. Quick access to Bernal Heights Park, 101/280, and 13 mins to SFO. Village charm with rapid Peninsula access.
- Listing Status:
- Active
- Date Added:
- April 30, 2026
- Data Last Updated:
- May 5, 2026 at 8:33PM
- Listing Office:
- Compass : 415-231-2312
- Listing Agent:
- Ruth Krishnan : 415 7355867
- MLS ID:
- 426126006

- Parking: CoveredGarage Facing FrontParking spaces: 2
- Windows: Skylight(s)
- Road/Access: Paved
- HOA Amenities: 0.0
- Originating MLS: San Francisco
- County: San Francisco
- Zoning: 5690-039
- View: CityHills
- Fencing: Wood
- Utilities: Public
- Water: Public
- Source: San Francisco
This listing courtesy of Ruth Krishnan , Compass
Monthly Payment
- Principal & Interest $
- Property Taxes $
- Home Insurance $
- VA Funding Fee $






