Canadian Shai Gilgeous Alexander, Serbian Nikola Jokic, Greek Giannis Antetokounmpo and Americans Donovan Mitchell and Jayson Tatum make up the NBA’s first team of the year announced Friday.
Oklahoma City Thunder’s Gilgeous Alexander, Denver Nuggets’ Jokic, Milwaukee Bucks’ Giannis and Boston Celtics’ Tatum were unanimously selected by the voting panel.
Three of the five members of the NBA’s first team are not American players, which marks an upward trend in the weight of foreign stars in the league.
The second team included Jalen Brunson of the New York Knicks, Steph Curry of the Golden State Warriors, Anthony Edwards of the Minnesota Timberwolves, LeBron James of the Los Angeles Lakers and Evan Mobley of the Cavaliers.
The third team included Cade Cunningham of the Detroit Pistons, Tyrese Haliburton of the Indiana Pacers, James Harden of the Los Angeles Clippers, Karl Anthony Towns of the New York Knicks and Jalen Williams of the Oklahoma City Thunder.
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