The Cleveland Cavaliers and Los Angeles Clippers don't have a whole lot in common. With the Cavs being the East's reiging champs the last 3 years, the Clippers have consistently put themselves on the other side of the spectrum, being submerged in mediocrity for the same amoung of time.
Despite these differences, however, the two teams may come together on a trade deal that would benefit both sides -- for different reasons.
According to Terry Pluto of the Plain Dealer, the Clippers would strongly consider a deal with the Cavs... if the Brooklyn 2018 first-round draft pick (that they acquired in the Kyrie Irving deal) was included. The proposed deal would reportedly send DeAndre Jordan to the Cavaliers for Tristan Thompson, Channing Frye, and the 2018 Nets pick.
Interestingly enough, Pluto adds, that 2018 pick is not something the Cavs are willing to give up just yet.
It's long been known that the likely lottery pick will serve as a "Plan B" for Cleveland if LeBron James elects to leave town. And with The King reportedly unwilling to commit long-term, it doesn't appear as if they'll change that plan anytime soon. With this in mind, it sounds unlikely that this deal will get done. But it does say a few important things about both teams:
For one, it reveals that the Clippers are at least entertaining the idea of shipping out DeAndre Jordan, which means they are at least considering a possible rebuild.
Secondly, it proves that the Cavaliers are unconfident in their ability to sign LeBron James this summer. Their unwillingness to trade a lottery pick for an All-Star center is something uncharacteristic of a stable contending team (unless you're Danny Ainge and the Boston Celtics).
No matter which way you look at it, it's clear that these franchises are looking to make some changes as the deadline draws near. What those changes will look like is anyone's guess, but if it's even close to the magnitude of this deal, it could shake up what's to come in April, May, and June.