For an empty tree, the number of nodes is $0$.
For a non-empty tree :
$\text{total nodes = 1 + nodes in left subtree + nodes in right subtree}$
Only option A follows this logic correctly :
Base Case
$\texttt{if t == NULL return 0;}$
If the pointer reaches a dead end (an empty subtree), it contributes $0$ to the total node count.
Recursive Step
$\texttt{return 1 + tree_size(t->left) + tree_size(t->right);}$
If the node exists, it counts itself as $1$, and then recursively adds the total number of nodes in its entire left subtree and its entire right subtree.
B, C, and D attempt to use an accumulator variable $(\texttt{count})$, but they all implement the logic incorrectly :
B : Never actually increments $\texttt{count}$ or adds $1$ for the current node.
It just passes the same $\texttt{count}$ value down and adds the results, which will yield an incorrect sum.
C : Increments $\texttt{count}$ for both the left and right recursive calls and then adds them together.
If you pass a tree with exactly $1$ node $($and an initial count of $0)$, it will $\texttt{return (0 + 1) + (0 + 1) = 2}$, which is wrong.
D : Asymmetrically increments the count (only adding to the right subtree), which makes no logical sense for calculating the total size of a tree.
