Consider the pseudocode given below. The function DoSomething() takes as argument a pointer to the root of an arbitrary tree represented by the leftMostChild-rightSibling representation. Each node of the tree is of type treeNode.typedef struct treeNode* treeptr; struct treeNode { treeptr leftMostChild, rightSibling; }; int DoSomething (treeptr tree) { int value=0; if (tree != NULL) { if (tree->leftMostChild == NULL) value = 1; else value = DoSomething(tree->leftMostChild); value = value + DoSomething(tree->rightSibling); } return(value); }When the pointer to the root of a tree is passed as the argument to DoSomething, the value returned by the function corresponds to the

In the leftMostChild-rightSibling representation of an arbitrary tree:

1. A node's leftMostChild pointer points to its first child in the original tree.

2. A node's rightSibling pointer points to the immediate next sibling among the children of the same parent.

A node in the arbitrary tree is a leaf node if and only if it has no children, which means tree->leftMostChild == NULL.

Let us trace the function DoSomething(tree):

• If tree == NULL, it returns 0.

• If tree != NULL:

  • If tree->leftMostChild == NULL, the current node has no children (it is a leaf node in the original arbitrary tree), so value is initialized to 1.

  • Otherwise, value recursively computes the number of leaves in the subtree rooted at its children by calling DoSomething(tree->leftMostChild).

  • Then, it adds DoSomething(tree->rightSibling) to recursively count leaves in all its siblings.

Thus, the function traverses the entire tree and counts every node that has leftMostChild == NULL, which is the total number of leaf nodes in the tree.

Hence, option (D) is correct.