Handle the case where there are one or two nodes on the binary space partition tree

The code assumed the tree has to have one node on one side and the two on the others after reaching this branch.

https://github.com/cshung/RotatingCube/blob/c4920354fc1809a061f14337d140515f97068675/RotatingCube/BinarySpacePartitionTreeNode.cs#L76

This might not be always true if nodes could be on the plane.