Boston Red Sox's Mookie Betts signals from second base after hitting a double against the Cleveland Indians during the sixth inning in Game 3 of baseball's American League Division Series, Monday, Oct. 10, 2016, in Boston. (AP Photo/Charles Krupa)
Boston Red Sox's Mookie Betts signals from second base after hitting a double against the Cleveland Indians during the sixth inning in Game 3 of baseball's American League Division Series, Monday, Oct. 10, 2016, in Boston. (AP Photo/Charles Krupa) Credit: Charles Krupa

NEW YORK – Boston Red Sox outfielder Mookie Betts has won his first Gold Glove award.

The honors for fielding excellence were announced Tuesday by Rawlings, with Betts being chosen as the American League’s top defensive right fielder.

Betts committed just one error in 157 games this past season, tallying 346 put outs and 14 assists.

Young shortstop Francisco Lindor from AL champion Cleveland was a first-time winner, along with Texas first baseman Mitch Moreland, Detroit second baseman Ian Kinsler, New York Yankees left fielder Brett Gardner.

Kansas City catcher Salvador Perez won his fourth Gold Glove in a row, Houston pitcher Dallas Keuchel took his third straight and Tampa Bay center fielder Kevin Kiermaier repeated from last year.

Rangers third baseman Adrian Beltre was rewarded for the fifth time overall.

First baseman Anthony Rizzo and right fielder Jason Heyward from the World Series champion Chicago Cubs won awards in the NL.