Question

Let x be an internal node in a binary search tree and y its successor in the infix tree-traversal (then x.key cannot be maximal).

Show that either x.right = null and y.left ≠ null, OR x.right ≠ null and y.left = null. (if provide pseudo-code, use Python)

Answer 1

ANS:---

# Python Programm for INFIX TREE-TRAVERSAL without recursion

  

# A binary tree node

class Node:

  

# Constructor to create a new node

def _init_(self, data):

self.data = data

self.left = None

self.right = None

  

# Iterative function for inorder tree traversal

def inOrder(root):

  

# Set current to root of binary tree

current = root

stack = [] # initialize stack

done = 0

  

while True:

  

# Reach the left most Node of the current Node

if current is not None:

  

# Place pointer to a tree node on the stack

# before traversing the node's left subtree

stack.append(current)

  

current = current.left

  

  

# BackTrack from the empty subtree and visit the Node

# at the top of the stack; however, if the stack is

# empty you are done

elif(stack):

current = stack.pop()

print(current.data, end=" ") # Python 3 printing

  

# We have visited the node and its left

# subtree. Now, it's right subtree's turn

current = current.right

  

else:

break

print()

  

# Driver program to test above function

  

""" Constructed binary tree is

1

/ \

2 3

/ \

4 5 """

  

root = Node(1)

root.left = Node(2)

root.right = Node(3)

root.left.left = Node(4)

root.left.right = Node(5)

inOrder(root)

NOTE:-Please check the code. If you have any doubts comment below and I am happy to help