Charles Barkley leaves Kobe and LeBron out of his list of top 5 NBA players of all-time

According to Charles Barkley, Kobe Bryant and LeBron James aren’t among the five best NBA players of all-time.

Barkley revealed his list of five best in a conversation with Kentucky coach John Calipari this week. At the top, unsurprisingly, was Michael Jordan.

“Michael’s one, Oscar Robertson’s two,” Barkley said. “[Bill] Russell, Wilt [Chamberlain] and Kareem [Abdul-Jabbar], no particular order after Michael. Kobe six, LeBron seven, then you’ve got Elgin Baylor, Jerry West.”

That’s a list that has some heavy throwback flavor. And while you may argue with the order of Barkley’s top seven, you can’t quibble much with the inclusion of all seven of those names. Each of them is one of the greatest players in NBA history.

Barkley also explained why he had James (and, indirectly, Bryant) outside the top five. And if you’ve been watching ESPN’s “The Last Dance,” that reason is a familiar one.

“You know, I love LeBron and everything about him but I still think that the way they play the game today he didn’t want any part of those bad boy Pistons,” Barkley said. “Let me tell you something, those guys were out there trying to hurt people. I used to always tell people when you were playing the Pistons you had to call home and tell your family you love them just in case you never saw them again.”

We’re guessing Bryant was ahead of James on Barkley’s list in part because of Bryant’s number of championships. Bryant has five championships, two more than James. And we’ve yet to find out if James will get a chance to chase his fourth championship during a resumption of the 2019-20 season or if the quest for No. 4 will have to wait until the 2020-21 season.

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Nick Bromberg is a writer for Yahoo Sports.

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