# Answered Essay: Extend your python program below to include the recursive methods, as d

Extend your python program below to include the recursive methods, as described in part A , B ad C.

class BST:

def __init__(self,data):

self.root = data

self.left = None

self.right = None

def insert(self,data):

if self.root == None:

self.root = BST(data)

elif data > self.root:

if self.right == None:

self.right = BST(data)

else:

self.right.insert(data)

elif data < self.root:

if self.left == None:

self.left = BST(data)

else:

self.left.insert(data)

def inordertraversal(self):

if self.left != None:

self.left.inordertraversal()

print (self.root),

if self.right != None:

self.right.inordertraversal()

t = BST(9)

t.insert(8)

t.insert(3)

t.insert(6)

t.insert(12)

t.insert(10)

t.insert(11)

t.inordertraversal()

(a) numberOfNodes Accepts a BST node r and returns an integer value of the

number of nodes in the BST whose root is r.

(b) treeHeight Accepts a BST node r and returns and integer value of the height

of the BST whose root is r.

(c) numberOfLeaves Accepts a BST node r and returns an integer value of the

number of leaves in the BST whose root is r.

Comment: You may choose to submit pseudocode only for Question 9. Include it

along with your Part A. Each algorithm will require very few lines of code. Do not

use extra local (or global) variables, they are unnecessary. Think: if at the base

case, return base case solution” else return something involving a recursive call”.

#Python pseudocode

#(a) numberOfNodes accept a BST node r and return

#an integer value of number of nodes in the BST whose root is r

def numberOfNodes(r):

counter=1

if r == None:

return 0

else:

counter +=numberOfNodes(r.left)

counter +=numberOfNodes(r.right)

return counter

#(b) treeHeight Accepts a BST node r and returns and integer value of the height

#of the BST whose root is r.

def treeHeight(r):

if r == None:

return 0

lheight = treeHeight(r.left)

rheight = treeHeight(r.right)

if (lheight > rheight):

return lheight + 1

else:

return rheight + 1

#(c) numberOfLeaves Accepts a BST node r and returns an integer value of the

#number of leaves in the BST whose root is r.

def numberOfLeaves(r):

if r == None:

return 0

if (r.left == None and r.right == None):

return 1

else:

return numberOfLeaves(r.left) + numberOfLeaves(r.right)

class BST:

def __init__(self,data):

self.root = data

self.left = None

self.right = None

def insert(self,data):

if self.root == None:

self.root = BST(data)

elif data > self.root:

if self.right == None:

self.right = BST(data)

else:

self.right.insert(data)

elif data < self.root:

if self.left == None:

self.left = BST(data)

else:

self.left.insert(data)

def inordertraversal(self):

if self.left != None:

self.left.inordertraversal()

print (self.root),

if self.right != None:

self.right.inordertraversal()

t = BST(9)

t.insert(8)

t.insert(3)

t.insert(6)

t.insert(12)

t.insert(10)

t.insert(11)

t.inordertraversal()

print (‘nNumber Of Nodes:’ + str(numberOfNodes(t)))

print (‘nTree Height:’ + str(treeHeight(t)))

print (‘nNumber Of Leaves:’ + str(numberOfLeaves(t)))

Steps to run above Python program & output is shown in the screenshot.

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