NBA Rumors: Knicks won’t trade young core for Eric Bledsoe

NEW YORK, NY - OCTOBER 27: Frank Ntilikina
NEW YORK, NY - OCTOBER 27: Frank Ntilikina /
facebooktwitterreddit

NBA Rumors: New York Knicks will not trade any of their young pieces for Eric Bledsoe

With Eric Bledsoe on the trade block, the speculation to where he will ultimately will land is quite ridiculous. But it always is in the NBA. And as is always the case, when a big name hits the block, the New York Knicks have surfaced as a possible landing spot.

While New York would most definitely welcome a player like Bledsoe, they have made it clear that they will not trade any of their young pieces for Eric Bledsoe.

And that’s probably the smart play here, especially for a franchise that’s looking for stability.

More from Sir Charles In Charge

"[via New York Post]In a State of the Knicks address on Sunday at Quicken Loans Arena, Knicks president Steve Mills said Ntilikina and Hernangomez are keepers."

After a summer in which the Knicks essentially cleaned house. First, by moving on past Phil Jackson and then second by finally trading Carmelo Anthony, the Knicks would be ridiculed if they did all that only to go around and move a big piece of their future for a soon to be 28-year-old that may have already be past his prime (due to injury).

Bledsoe is a fine player. In fact, he’s arguably a top 10-15 point guard when he’s healthy and on top of his game. With that said, he’s not a franchise-changing player that is worth mortgaging your entire – or even partial – young core for.

The New York Knicks don’t need to look for the quick fix. If they want to get back to a respectable place in the NBA again, they need to trust their young pieces. It’s going to take a while, but the long rebuild is always the safest bet.

Must Read: Milwaukee Bucks should keep Malcolm Brogdon, trade Jabari Parker

And if a big free agent comes along the way, perhaps this or next summer, that’s just gravy. No need to force the issue. Not right now. Not for this team. They’ll be OK in the long run.