Sources: Dwyane Wade agrees to one-year, $2.3M deal to join Cavs

Shams Charania
The Vertical
Dwyane Wade is a 12-time All-Star and three-time NBA champion. (AP)

Dwyane Wade has agreed to a one-year, $2.3 million deal to join the Cleveland Cavaliers, league sources told The Vertical.

Wade reached a contract buyout with the Chicago Bulls on Sunday and is expected to sign the deal with the Cavs this week after clearing waivers on Wednesday.

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For Wade, the lure of reuniting with LeBron James and competing for a championship in the Eastern Conference outweighed suitors such as Oklahoma City, San Antonio and Miami. Wade is expected to give additional playmaking and scoring to the Cavaliers, who must now waive a guaranteed contract by the start of the regular season.

Wade, 35, played four seasons with James in Miami, where the pair won two NBA championships.

Wade left the Heat last summer for a two-year, $47 million contract with the Bulls. He lost about $8 million in negotiating his release from Chicago, but now joins a team that has been to three straight NBA Finals, winning the league championship in 2016.

Wade averaged 18.3 points, 4.5 rebounds and 3.8 assist and shot 43.3 percent from the floor last season with Chicago. In 14 NBA seasons, the 2003 No. 5 overall pick from Marquette has averaged 23.3 points, 5.7 assists and 4.8 rebounds while shooting 48.4 percent from the floor.

Wade spent his first 13 seasons with the Heat, leading Miami to the NBA title in 2006.

Wade is a 12-time All-Star, three-time NBA champion, 2009 scoring champion and eight-time All-NBA player.

Cleveland plans to waive rookie JaCorey Williams to clear a roster spot for Wade, league sources told The Vertical. Williams plans to find his next team soon.

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