In A binary tree if there are $N$ leaf nodes then the number of nodes having 2 children will be N-1
$\text{Proof}:$
Key idea is find number of edges using Degree and find number of edges using nodes and equate them
Let $l$ be the number of leaf nodes, $D_1$ be the number of nodes with one child and $D_2$ be the number of nodes with two children.
$\text{Sum of Degrees}, D = 1 \times l + 3\times D_2 + 2\times D_1 - 1 \text{(for root)}$
(Root is having one degree less because it is not having a parent)
$D= l+ 3D_2+ 2D_1 - 1 \qquad \rightarrow (1)$
$\text{Number of edges}, e= \frac{D}{2} $
$\text{Number of nodes}, n = D_2+D_1+l $
A tree with $n$ nodes has $n-1$ edges so,
$\frac{D}{2} = D_2+D_1+l-1\qquad \to(2)$
From $(1)$ and $(2)$
$\frac{l+3D_2+2D_1-1}{2} = D_2+D_1+l-1$
$\implies D_2 = l-1$
So, the number of nodes in $T$ having two children $= 20-1 = 19$