By seeing the degree's of nodes, i understood TREE assumed as Directed graph.
Let N$_i$ represent Number of Nodes with Degree i
Total No.of Nodes in a Tree = N$_0$ + N$_1$ + N$_2$ + N$_3$
In a tree with N nodes have exactly n-1 edges.
∴ N = |E| + 1 ==> N$_0$ + N$_1$ + N$_2$ + N$_3$ = |E| + 1 --------- (1)
Given that, N$_1$ = 4 , N$_2$ = 3 , N$_3$ = 3
Total Edges in the directed graph = summation of all degrees = ∑ N$_i$ . i
∴ | E | = ( 0*N$_0$) + (1*N$_0$) + (2*N$_2$) + (3*N$_3$) = ( 0*$\color{red}?$) + (1*4) + (2*3) + (3*3) = 0+4+6+9 = 19
Substitute this value in eqn (1),
N$_0$ + N$_1$ + N$_2$ + N$_3$ = 19 + 1
N$_0$ + 4+3+3 = 19+1 ===> N$_0$ + 10 = 20 ===> N$_0$ = 10