Morris traversal for Preorder
Using Morris Traversal, we can traverse the tree without using stack and recursion.
Morris traversal for Preorder
1…If left child is null, print the current node data. Move to right child.
….Else, Make the right child of the inorder predecessor point to the current node. Two cases arise:
………a) The right child of the inorder predecessor already points to the current node. Set right child to NULL. Move to right child of current node.
………b) The right child is NULL. Set it to current node. Print current node’s data and move to left child of current node.
2…Iterate until current node is not NULL
Following is the implementation of the above algorithm.
# Python program for Morris Preorder traversal
# A binary tree Node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Preorder traversal without
# recursion and without stack
def MorrisTraversal(root):
curr = root
while curr:
# If left child is null, print the
# current node data. And, update
# the current pointer to right child.
if curr.left is None:
print(curr.data, end= " ")
curr = curr.right
else:
# Find the inorder predecessor
prev = curr.left
while prev.right is not None and prev.right is not curr:
prev = prev.right
# If the right child of inorder
# predecessor already points to
# the current node, update the
# current with it's right child
if prev.right is curr:
prev.right = None
curr = curr.right
# else If right child doesn't point
# to the current node, then print this
# node's data and update the right child
# pointer with the current node and update
# the current with it's left child
else:
print (curr.data, end=" ")
prev.right = curr
curr = curr.left
# Function for sStandard preorder traversal
def preorfer(root):
if root :
print(root.data, end = " ")
preorfer(root.left)
preorfer(root.right)
# Driver program to test
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left= Node(6)
root.right.right = Node(7)
root.left.left.left = Node(8)
root.left.left.right = Node(9)
root.left.right.left = Node(10)
root.left.right.right = Node(11)
MorrisTraversal(root)
print("\n")
preorfer(root)
# This code is contributed by 'Aartee'
Limitations:
Morris traversal modifies the tree during the process. It establishes the right links while moving down the tree and resets the right links while moving up the tree. So the algorithm cannot be applied if write operations are not allowed.
References:
https://www.geeksforgeeks.org/morris-traversal-for-preorder/