Short answer: yes
Long answer: Consider a rectangle in 2D (works the same, but has less sides).
Every node splits the space into two halves. Leaf nodes labels it's space as empty or solid.
So for a rectangle
++
 
 
 
++
you could start by choosing the left side as the first splitting plane (in 2D a splitting line):
A  B


++
 
 
 
++


So your root node now has a splitting plane and two children A and B. A will remain a leaf node and label the left hand side as empty.
B we could now split further by, for example, the top edge:
A 

 [B] C
++
  [B] D
 
 
++


so the node B now has two children C and D. C again labeling it's space as empty. Now after splitting D for example by the right hand side we get:
A 

 [B] C
++
 
 
 
++
 
 
 
[B,D] F[B,D] E
and there you have it: one splitting plane on the left, and further down the tree one on the right. E, which is a child node of D, which again is a child node of B, denotes the right hand side as empty. Now we can finish things up by splitting F
A 

 [b] C
++
 
 
B,D,F]H
++
B,D,F]G
 
 
 [B,D] E
and labeling G as empty and H as solid.
And there you have it, a BSP tree with a total of 4 splits, one for each side.