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, rightSibiling; 
}; 
Int Dosomething (treeptr tree) 
{ 
    int value =0; 
    if (tree ! = NULL) { 
      If (tree -> leftMostchild = = NULL) 
          value=1; 
    else 
        value = Dosomething (tree->leftMostchild); 
        value = value + Dosometing (tree->rightsibiling); 
    } 
    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

Suppose we have a balanced binary search tree T holding n numbers. We are given two numbers L and H and wish to sum up all the numbers in T that lie between L and H. Suppose there are m such numbers in T. If the tightest upper bound on the time to compute the sum is O( n^a log^b n + m^c log^d n ), the value of a + 10b + 100c + 1000d is _______.

Consider the following rooted tree with the vertex labelled P as the root The order in which the nodes are visited during an in-order traversal of the tree is

Consider the following relational schema:

employee (empId, empName, empDept )

customer (custId, custName, salesRepId, rating)

SalesRepId is a foreign key referring to empId of the employee relation. Assume that each employee makes a sale to at least one customer. What does the following query return?

SELECT empName 
FROM employee E 
WHERE NOT EXISTS (SELECT custId 
       FROM customer C 
       WHERE C.salesRepId = E.empId 
       AND C.rating <> 'GOOD');