Answer:

A recursive subroutine is simply one that calls itself either directly or through a chain of calls involving other subroutines.

Answer:

The outputs are:

printStuff(0) outputs:   *
printStuff(1) outputs:   [*,*]
printStuff(2) outputs:   [[*,*],[*,*]]
printStuff(3) outputs:   [[[*,*],[*,*]],[[*,*],[*,*]]]

(Explanation: For printStuff(0), the value of the parameter is 0, so the first clause of the if is executed, and the output is just *. For printStuff(1), the else clause is executed. This else clause contains two recursive calls to printStuff(level-1). Since level is 1, level-1 is 0, so each call to printStuff outputs a *. The overall output from printStuff(1) is [*,*]. In a similar way, printStuff(2) includes two recursive calls to printStuff(1). Each call to printStuff(1) outputs [*,*]. And printStuff(2) just takes two copies of this and puts them between [ and ] separated by a comma: [[*,*],[*,*]]. Finally, the output from printStuff(3) outputs two copies of [[*,*],[*,*]] separated by a comma and enclosed between brackets. Once you recognize the pattern, you can do printStuff(N) for any N without trying to follow the execution of the subroutine in detail.)

Answer:

I'll give both a non-recursive solution and a recursive solution. For a linked list, the recursion is not really necessary, but it does nicely reflect the recursive definition of ListNode

static int countZeros( ListNode head ) {
   int count;        // The number of zeros in the list.
   ListNode runner;  // For running along the list.
   count = 0;
   runner = head;    // Start at the beginning of the list.
   while (runner != null) {
      if ( runner.item == 0)
         count++;  // Count the zero found in the current node.
      runner = runner.next;  // Advance to the next node.
   }
   return count;
}

static int countZerosRecursively( ListNode head ) {
   if ( head == null) {
          // An empty list does not contain any zeros.
       return 0;
   }
   else {
       int count = countZerosRecursively( head.next );  // Count zeros in tail.
       if ( head.item == 0 )
           count++;  // Add 1 to account for the zero in the head node.
       return count;
   }
}

Answer:

ListNode tail;  // Since no tail pointer is given, we need to make one
tail = head;  // We are assuming that this is not null.
while (tail.next != null) {
    tail = tail.next;
}
// At this point, tail points to the last node in the list
ListNode node42;
node42 = new ListNode();  // create a new node
node42.item = 42;    // the item in the new node is 42
tail.next = node42;  // attach the new node to the list

(Explanation: To add a list node at the end of the list, we need a pointer to the last node in the list. Since no such pointer is given, we have to make one. Start with a pointer that is equal to head, and move it down the list until it points to the last node in the list. We know that tail is pointing to the last node in the list if tail.next is null. Use a while loop to move tail down the list until that is true. Then we can make a new node and attach it to the list by setting tail.next to point to it. The value of node42.next is null, and that null marks the new end of the list.)

Answer:

The three stack operations are push, pop, and isEmpty. The definitions of these operations are: push(item) adds the specified item to the top of the stack; pop() removes the top item of the stack and returns it; and isEmpty() is a boolean-valued function that returns true if there are no items on the stack.

Answer:

In a stack, items are added to the stack and removed from the stack on the same end (called the "top" of the stack). In a queue, items are added at one end (the "back") and removed at the other end (the "front"). Because of this difference, a queue is a FIFO structure (items are removed in the same order in which they were added), and a stack is a LIFO structure (the item that is popped from a stack is the one that was added most recently).

Answer:

When a subroutine is called, an activation record is created to hold the information that is needed for the execution of the subroutine, such as the values of the parameters and local variables. This activation record is stored on a stack of activation records. A stack is used since one subroutine can call another, which can then call a third, and so on. Because of this, many activation records can be in use at the same time. The data structure is a stack because an activation record has to continue to exist while all the subroutines that are called by the subroutine are executed. While they are being executed, the stack of activation records can grow and shrink as subroutines are called and return.

Answer:

static int treeSum( TreeNode root ) {
       // Find the sum of all the nodes in the tree to which root points.
    if ( root == null ) {
          // The sum of the nodes in an empty tree is zero.
       return 0;
    }
    else {
          // Add the item in the root to the sum of the
          // items in the left subtree and the sum of the
          // items in the right subtree.
       int total = root.item;
       total += treeSum( root.left );
       total += treeSum( root.right );
       return total;
    }
 }

Answer:

static TreeNode copyTree( TreeNode root ){
        // Make a copy of the tree that root points to.
    if (root == null) {
        return null;  // The copy of an empty tree is an empty tree
    }
    else {
            // The tree is not empty.  We need to make a new node
            // to be the root of the copy, and then we need to copy
            // the left and right subtrees.
         TreeNode rootOfCopy = new TreeNode();
         rootOfCopy.item = root.item;
         rootOfCopy.left = copyTree( root.left );
         rootOfCopy.right = copyTree( root.right );
         return rootOfCopy;
    }
}

(The essential point here is that the subtrees of the original tree can be copied using recursive calls to copyTree(). For example, copyTree(root.left) will make a complete copy of the left subtree of the original tree, and that can become the left subtree of the new root node. This works even if root.left is null.)

Answer:

In a traversal of a binary tree, all the nodes are processed in some way. (For example, they might be printed.) In a postorder traversal, the order of processing is defined by the rule: For each node, the nodes in the left subtree of that node are processed first. Then the nodes in the right subtree are processed. Finally, the node itself is processed. This rule is applied at all levels of the tree.

Answer:

Here are five possibilities (out of an infinite number of possibilities), with some explanation:

fred — A <multilist> can just be a word, such as "fred".
          
( )  — The [ ]... around <multilist> means that there can be
          any number of nested <multilist>'s, including zero.  If
          there are zero, then all that's left is the empty
          parentheses.
          
( fred mary chicago )—A <multilist> consisting of three
                         <multilist>'s—"fred", "mary", and
                         "chicago"—inside parentheses
                         
( ( able ) ( baker charlie ) )—A <multilist> containing two
                                  <multilist>'s.
                                  
( ( a ( b ) ) ( c ( d e ) g ) ) —Even more nesting.

Answer:

To parse a computer program means to determine its syntactic structure, that is, to figure out how it can be constructed using the rules of a grammar (such as a BNF grammar).