DecryptNode for non-leaf Node in CP-ABE
The last two steps is from Lagrange interpolation such that $\sum_{i\in S_x'} q_x(i)\triangle_{i,S_x'}(0)=q_x(0)$. The concept is the same as that of leaf node, but now we are at the parent node. Doing this recursively up to the root node would give the secret exponent.