We have $4$ nodes out of which $1$ is a leaf and there are no mode leaf nodes. So, we must have $4$ levels with one node in each. Each node in levels $2$, $3$ and $4$ have two choices- either to be left child or right child. So, totally $2 \times 2 \times 2 = 8$ ways. Now, $2,3,4$ can be permuted in any order in first $3$ levels and each permutation gives a different binary tree. So, total number of binary trees $= 8 \times 3! = 48$.