WARSAW, Poland -- Alexandra Dulgheru, a Romanian qualifier ranked No. 201, won the Warsaw Open on Saturday in her WTA Tour debut by beating Alona Bondarenko of Ukraine 7-6 (3), 3-6, 6-0.
In a match played in strong wind and twice interrupted by rain, the 19-year-old Dulgheru used an array of groundstrokes and drop shots to break her eighth-seeded opponent eight times. Bondarenko, ranked No. 39, was the runner-up in this tournament in 2007.
"My first plan was to try and pass the [qualifications] of the tournament, and doing that was a great achievement for me," Dulgheru said. "I did not expect this and I'm really happy that I won."
Bondarenko had two chances to serve out the first set but couldn't capitalize, with Dulgheru breaking to force a tiebreaker. Bondarenko rallied with two breaks in the second to lead 4-0 before the match was delayed about 25 minutes because of rain.
Dulgheru broke serve to lead 2-0 in the final set before rain again sent the players from the court. After play resumed about 10 minutes later, Dulgheru reeled off four straight games to close the match.
"After every game I was feeling more and more tired, so I'm very happy that I managed to pull through in the end," Dulgheru said.
Dulgheru did not qualify for the French Open. Bondarenko takes on 20th-seeded Dominika Cibulkova of Slovakia in the first round at Roland Garros.