CS313E: Fall 2011 -- Quiz5 Answers

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1. (10 points) Assume you have a binary expression tree representing
  (5 + 8) \times (7 + 2) and that ``visiting'' a node means printing
  the value stored at that node:


The tree will be:

             *
           /   \
          +     +
         / \   / \
        5   8 7   2

a. What is printed for an inorder traversal of the tree?

   5 + 8 * 7 + 2

b.  For a postorder traversal?

   5 8 + 7 2 + *

c.  For a preorder traversal?

   * + 5 8 + 7 2

d.  For a level order traversal? 

   * + + 5 8 7 2

e. What is the complexity of each of these traverals?

   They are all linear. 


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2. (5 points) Write a Boolean-valued Python function  isInt( str ) 
  that will return True if a string represents a legal integer
  value (including negatives).  The string can have whitespace at
  either end (but not in the middle).  For example, isInt(str)
  should return True if str is any of " -12", "0 "
  or " 156 ", but return False on " 15 6 " or "- 3".

def isInt( str ):
        s = str.strip()
        return ( s.isdigit() or 
		 ( s[0] == '-' and s[1:].isdigit() ))

>>> from quiz5 import *
>>> isInt("-")
''
>>> isInt("-2")
True
>>> isInt("-2A")
False
>>> isInt("-123 45")
False

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3. (5 points) Write an integer valued Python function 
  countNodes( tree ), that will return the number of nodes in a BinaryTree.
  Remember that BinaryTree is really a ``wrapper'' around the
  BinaryTreeNode class, so you will probably need an auxiliary
  function.


def countNodesAux(node):
    if not node:
        return 0
    else:
        return 1 \
            + countNodesAux( node.getLeft() ) \
            + countNodesAux( node.getRight() ) 

def countNodes( tree ):
    root = tree.getRoot()
    if not root:
        return 0
    else:
        return countNodesAux( root )



import random
input = [ random.randint(0, 999) for i in range(200) ]
tree = BinarySearchTree()    
for i in input:
    tree.insert(i)
print("Tree contains ", countNodes( tree ), " nodes!")



def countNodesAux(node):
    if not node:
        return 0
    else:
        return 1 \
            + countNodesAux( node.getLeft() ) \
            + countNodesAux( node.getRight() ) 

def countNodes( tree ):
    root = tree.getRoot()
    if not root:
        return 0
    else:
        return countNodesAux( root )

# Some code to test it:

import random
input = [ random.randint(0, 999) for i in range(200) ]
tree = BinarySearchTree()    
for i in input:
    tree.insert(i)
print("Tree contains ", countNodes( tree ), " nodes!")

>>> from Trees import *
Tree contains 200 nodes!