Unit 9 Practice Test (AB27-30)

Transcription

1 Unit 9 Practice Test (AB27-30) Name 1. Consider the following method: public static int checktree(treenode root) return 0; int x = checktree(root.getleft()); if ( x >= 0 && checktree(root.getright()) == x) return x + 1; return 1; checktree(root) returns a non-negative value if and only if the tree rooted in root A) is not empty and the root is a leaf B) is empty or the root is a leaf or the root has two children C) has no nodes with only one child D) is empty or the root has left and right subtrees of the same depth E) is a full tree (it is balanced and every parent has two children) 2. Consider the following class. Assume any used methods of the IncomingCall objects are defined as intuitively make sense. public class CallerId private LinkedList calls; public CallerId() calls = new LinkedList(); //precondition: calls holds IncomingCall objects //postcondition: all calls that came from target are removed from the calls list public void removeall(string target) < code not shown > < other methods not shown > If removeall works as specified in the pre- and postconditions, which of the following code segments can serve as removeall s body? I. for (int i = 0; i < calls.size(); i++) if (((IncomingCall)calls.get(i)).getSource().equals(target)) calls.remove(i);

2 II. int i = 0; while (i < calls.size()) if (((IncomingCall)calls.get(i)).getSource().equals(target)) calls.remove(i); i++; III. Iterator iter = calls.iterator(); while (iter.hasnext()) if (((IncomingCall)iter.next()).getSource().equals(target)) iter.remove(); A) I only B) II only C) I and II D) II and III E) I, II, and III 3. Suppose ListNode head refers to the first node of a linked list. Consider the following code fragment: ListNode node, temp; for (node = head; node!= null; node = node.getnext()) while (node.getnext()!= null && node.getnext().getvalue().equals(node.getvalue())) node.setnext((node.getnext().getnext())); If head refers to a linked list with 11 nodes that hold the letters M, I, S, S, I, S, S, I, P, P, I (in that order), what letters are stored in the nodes of this list after the above code is executed? A) M B) M, I, S C) M, I, I, I, I D) M, I, S, P E) M, I, S, I, S, I, P, I

3 4. Consider the following method: public int mysterycount(treenode root) int count = 0; if (root!= null) count = mysterycount(root.getleft()) + mysterycount(root.getright()); if ((root.getleft() == null && root.getright() == null) (root.getleft()!= null && root.getright()!= null && root.getleft().getvalue().equals( root.getright().getvalue()))) count++; return count; If root refers to the tree which value is returned by mystercount(root)? A) 1 B) 3 C) 4 D) 5 E) 10 A / \ / \ B C / \ / \ D D C C / \ / \ E E E E 5. a) Write a boolean method isvalidautree as started below. The method returns true if and only if the tree is empty or if it is a valid expression tree: that is, the children of each node match the number of inputs for the AU stored in that node. //precondition: root is null or refers to the root of a binary // tree that hods AU objects as values in its nodes. //postcondition: returns true if root is null or if root refers to // a valid AU tree; otherwise returns false; in a // valid AU tree the number of children of each node // matches the number of inputs of the AU stored in // that node; a node with a one-input AU has only a // left child. Solution 1 public static boolean isvalidautree(treenode root) if((root.getvalue().numinputs()==0)&& (root.getleft()==null)&&

4 (root.getright()==null))) if ((root.getvalue().numinputs()==1)&& (root.getleft().isvalidautree())&& (root.getright()==null))) if ((root.getvalue().numinputs()==2)&& (root.getleft().isvalidautree())&& (root.getright().isvalidautree())) return false; Solution 2 public static boolean isvalidautree(treenode root) int numchildren; TreeNode left = root.getleft(); TreeNode right = root.getright(); if (left == null && right!= null) return false; if (left!= null && right!= null) numchildren = 2; if (left!= null && right == null) numchildren = 1; numchildren = 0; return numchildren == (root.getvalue().numinputs() && isvalidautree(left) && isvalidautree(right)) b) To evaluate an AU tree, we need to evaluate the left and right subtrees of the root, if present, then use the results as the inputs for the AU in the root. Write the method evaluate as started below. //precondition: root!= null; // isvalidautree(root) == true //postcondition: returns the value computed from the expression // tree root; the register field of the AU in each // node is set to the result of evaluating that // node s subtree public static int evaluate(treenode root)

5 TreeNode left = root.getleft(); TreeNode right = root.getright(); int x = 0, y = 0; if (left!= null) x = evaluate(left); if (right!= null) y = evaluate(right); return ((AU)root.getValue()).compute(x, y); 6. Review all vocabulary words from AB28-30.