Morris Traversal Inorder Without Stack in DSA Python - Build from Scratch

Morris Traversal Inorder Without Stack

📖 Scenario: You have a binary tree representing a family tree. You want to visit each family member in order without using extra memory like a stack or recursion.

🎯 Goal: Build a program that performs Morris inorder traversal on a binary tree and prints the nodes in sorted order without using stack or recursion.

📋 What You'll Learn

Create a binary tree with given nodes

Use Morris traversal technique to traverse inorder without stack

Print the nodes in inorder sequence

💡 Why This Matters

🌍 Real World

Morris traversal is useful when memory is limited and you want to traverse trees efficiently without extra space.

💼 Career

Understanding Morris traversal helps in optimizing tree traversals in embedded systems and performance-critical applications.

1

Create the Binary Tree Node Class and Tree

Create a class called Node with attributes val, left, and right. Then create the root node with value 4, and add left child 2 and right child 5. Also add children 1 and 3 to node 2.

DSA Python

# Define Node class and create the tree nodes
# Your code here

Hint

Define a class with __init__ method and create nodes by assigning left and right children.

2

Initialize Current Pointer for Traversal

Create a variable called current and set it to root. This will be used to traverse the tree.

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

root = Node(4)
root.left = Node(2)
root.right = Node(5)
root.left.left = Node(1)
root.left.right = Node(3)

# Set current to root for traversal
# Your code here

Hint

Just assign current to root to start traversal.

3

Implement Morris Inorder Traversal Logic

Write a while loop that runs while current is not None. Inside, if current.left is None, print current.val and move current to current.right. Otherwise, find the inorder predecessor of current by moving to the rightmost node in current.left. If predecessor's right is None, set it to current and move current to current.left. Else, set predecessor's right to None, print current.val, and move current to current.right.

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

root = Node(4)
root.left = Node(2)
root.right = Node(5)
root.left.left = Node(1)
root.left.right = Node(3)

current = root

# Implement Morris inorder traversal loop
# Your code here

Hint

Use a loop to traverse. Use the predecessor's right pointer to create and remove temporary links.

4

Print the Inorder Traversal Result

Run the program to print the inorder traversal of the tree using Morris traversal. The output should be the node values in sorted order separated by spaces.

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

root = Node(4)
root.left = Node(2)
root.right = Node(5)
root.left.left = Node(1)
root.left.right = Node(3)

current = root

while current is not None:
    if current.left is None:
        print(current.val, end=' ')
        current = current.right
    else:
        predecessor = current.left
        while predecessor.right is not None and predecessor.right != current:
            predecessor = predecessor.right
        if predecessor.right is None:
            predecessor.right = current
            current = current.left
        else:
            predecessor.right = None
            print(current.val, end=' ')
            current = current.right

# Run the program to see output
# Your code here

Hint

The output should show the values in sorted order: 1 2 3 4 5

Practice

1. What is the main advantage of Morris Traversal for inorder tree traversal compared to the traditional recursive or stack-based methods?

easy

A. It uses no extra space for stack or recursion.

B. It is faster than all other traversal methods.

C. It only works on binary search trees.

D. It modifies the tree permanently.

Solution

1. Step 1: Understand Morris Traversal space usage

   Morris Traversal uses threaded binary trees to avoid recursion or stack, so it uses O(1) extra space.

2. Step 2: Compare with other methods

   Traditional inorder traversal uses recursion or stack which requires O(h) space, where h is tree height.

3. Final Answer:

   It uses no extra space for stack or recursion. -> Option A

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Morris uses threads, no stack or recursion needed [OK]

Common Mistakes:

• Thinking Morris is faster in time complexity
• Believing it only works on BSTs
• Assuming it permanently changes the tree

2. Which of the following code snippets correctly initializes the current node for Morris inorder traversal in Python?

easy

A. current = root

B. current = None

C. current = root.left

D. current = root.right

Solution

1. Step 1: Identify starting point for traversal

   Morris traversal starts from the root node to cover the entire tree.

2. Step 2: Check initialization options

   Setting current = root ensures traversal begins at the tree's root.

3. Final Answer:

   current = root -> Option A

4. Quick Check:

   Start at root node [OK]

Hint: Always start traversal from root node [OK]

Common Mistakes:

• Starting from None causes no traversal
• Starting from root.left misses root node
• Starting from root.right misses left subtree

3. What will be the output of the Morris inorder traversal on the following binary tree?

    2
   / \
  1   3

Assuming the traversal prints node values in order.

medium

A. [2, 1, 3]

B. [1, 3, 2]

C. [3, 2, 1]

D. [1, 2, 3]

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits left subtree, root, then right subtree.

2. Step 2: Apply Morris traversal order

   For the tree, left is 1, root is 2, right is 3, so output is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option D

4. Quick Check:

   Inorder = left-root-right = [1, 2, 3] [OK]

Hint: Inorder = left, root, right always [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Misreading tree structure
• Assuming traversal prints in insertion order

4. Consider this snippet of Morris inorder traversal code in Python:

current = root
while current:
    if current.left is None:
        print(current.val)
        current = current.right
    else:
        pre = current.left
        while pre.right and pre.right != current:
            pre = pre.right
        if pre.right is None:
            pre.right = current
            current = current.left
        else:
            pre.right = None
            print(current.val)
            current = current.right

What is the main bug if the line pre.right = None is missing?

medium

A. The traversal will print nodes in preorder instead of inorder.

B. The traversal will skip the root node.

C. The traversal will enter an infinite loop.

D. The traversal will cause a syntax error.

Solution

1. Step 1: Understand role of pre.right = None

   This line removes the temporary thread created to return to the root after left subtree traversal.

2. Step 2: Consequence of missing this line

   If not removed, the thread remains, causing the loop to revisit nodes endlessly, creating an infinite loop.

3. Final Answer:

   The traversal will enter an infinite loop. -> Option C

4. Quick Check:

   Missing thread removal = infinite loop [OK]

Hint: Always remove temporary threads to avoid loops [OK]

Common Mistakes:

• Thinking it causes syntax error
• Assuming it changes traversal order
• Believing it skips nodes

5. You want to modify Morris inorder traversal to collect node values in a list instead of printing them. Which of the following code changes correctly achieves this without changing the traversal logic?

result = []
current = root
while current:
    if current.left is None:
        # Change here
        current = current.right
    else:
        pre = current.left
        while pre.right and pre.right != current:
            pre = pre.right
        if pre.right is None:
            pre.right = current
            current = current.left
        else:
            pre.right = None
            # Change here
            current = current.right

hard

A. Replace only second '# Change here' line with result.append(current.val)

B. Replace both '# Change here' lines with result.append(current.val)

C. Replace only first '# Change here' line with result.append(current.val)

D. Replace both '# Change here' lines with print(current.val)

Solution

1. Step 1: Identify where nodes are visited

   In Morris traversal, nodes are visited and printed in two places: when left is None and after removing thread.

2. Step 2: Replace print with appending to list

   To collect values, replace both print statements with result.append(current.val) to capture all visited nodes.

3. Final Answer:

   Replace both '# Change here' lines with result.append(current.val) -> Option B

4. Quick Check:

   Collect all visited nodes in list [OK]

Hint: Append values at both visit points to collect all nodes [OK]

Common Mistakes:

• Appending only once misses some nodes
• Leaving print statements causes unwanted output
• Changing traversal logic breaks correctness