Morris Traversal Inorder Without Stack in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Morris Inorder Traversal on a Simple Tree

What is the output of the Morris inorder traversal on the following binary tree?

Tree structure:
2
/ \
1 3

Code snippet uses Morris traversal to print node values in order.

DSA Javascript

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function morrisInorder(root) {
  let result = [];
  let current = root;
  while (current !== null) {
    if (current.left === null) {
      result.push(current.val);
      current = current.right;
    } else {
      let pre = current.left;
      while (pre.right !== null && pre.right !== current) {
        pre = pre.right;
      }
      if (pre.right === null) {
        pre.right = current;
        current = current.left;
      } else {
        pre.right = null;
        result.push(current.val);
        current = current.right;
      }
    }
  }
  return result;
}

const root = new Node(2);
root.left = new Node(1);
root.right = new Node(3);
console.log(morrisInorder(root));

A [3, 2, 1]

B [1, 2, 3]

C [1, 3, 2]

D [2, 1, 3]

💡 Hint

Morris traversal prints nodes in inorder sequence without recursion or stack.

Morris traversal visits nodes in inorder: left subtree, root, right subtree. For the tree 2 with left child 1 and right child 3, inorder is 1, 2, 3.

❓ Predict Output

intermediate

Output of Morris Inorder Traversal on a Left-Skewed Tree

What is the output of the Morris inorder traversal on this left-skewed binary tree?

Tree structure:
4
/
3
/
2
/
1

DSA Javascript

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function morrisInorder(root) {
  let result = [];
  let current = root;
  while (current !== null) {
    if (current.left === null) {
      result.push(current.val);
      current = current.right;
    } else {
      let pre = current.left;
      while (pre.right !== null && pre.right !== current) {
        pre = pre.right;
      }
      if (pre.right === null) {
        pre.right = current;
        current = current.left;
      } else {
        pre.right = null;
        result.push(current.val);
        current = current.right;
      }
    }
  }
  return result;
}

const root = new Node(4);
root.left = new Node(3);
root.left.left = new Node(2);
root.left.left.left = new Node(1);
console.log(morrisInorder(root));

A [1, 2, 3, 4]

B [4, 3, 2, 1]

C [1, 3, 2, 4]

D [2, 1, 3, 4]

💡 Hint

Inorder traversal visits leftmost nodes first.

The inorder traversal of a left-skewed tree visits nodes from smallest to largest: 1, 2, 3, 4.

🔧 Debug

advanced

Identify the Error in Morris Traversal Implementation

The following code attempts Morris inorder traversal but produces incorrect output. What is the bug causing this?

Code snippet:

function morrisInorder(root) {
  let result = [];
  let current = root;
  while (current !== null) {
    if (current.left === null) {
      result.push(current.val);
      current = current.left;  // Bug here
    } else {
      let pre = current.left;
      while (pre.right !== null && pre.right !== current) {
        pre = pre.right;
      }
      if (pre.right === null) {
        pre.right = current;
        current = current.left;
      } else {
        pre.right = null;
        result.push(current.val);
        current = current.right;
      }
    }
  }
  return result;
}

A The code is missing a base case for empty tree

B The inner while loop condition should check 'pre.left' instead of 'pre.right'

C Line 'current = current.left;' should be 'current = current.right;'

D The result array should be initialized inside the while loop

💡 Hint

When left child is null, traversal should move to right child.

When current.left is null, we must move to current.right to continue traversal. Moving to current.left causes infinite loop or wrong output.

🧠 Conceptual

advanced

Why Morris Traversal Uses Threaded Binary Tree Links

Why does Morris inorder traversal temporarily modify the tree by creating links (threads) from the rightmost node of the left subtree back to the current node?

A To avoid using stack or recursion by remembering the path back to the current node

B To permanently change the tree structure for faster future traversals

C To balance the tree dynamically during traversal

D To store node values temporarily for sorting after traversal

💡 Hint

Think about how Morris traversal avoids extra memory usage.

Morris traversal creates temporary links to remember where to return after visiting left subtree, avoiding stack or recursion.

❓ Predict Output

expert

Output of Morris Traversal on Complex Tree

What is the output of the Morris inorder traversal on this binary tree?

Tree structure:
5
/ \
3 7
/ \ \
2 4 8
/
1

DSA Javascript

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function morrisInorder(root) {
  let result = [];
  let current = root;
  while (current !== null) {
    if (current.left === null) {
      result.push(current.val);
      current = current.right;
    } else {
      let pre = current.left;
      while (pre.right !== null && pre.right !== current) {
        pre = pre.right;
      }
      if (pre.right === null) {
        pre.right = current;
        current = current.left;
      } else {
        pre.right = null;
        result.push(current.val);
        current = current.right;
      }
    }
  }
  return result;
}

const root = new Node(5);
root.left = new Node(3);
root.right = new Node(7);
root.left.left = new Node(2);
root.left.right = new Node(4);
root.right.right = new Node(8);
root.left.left.left = new Node(1);
console.log(morrisInorder(root));

A [1, 3, 2, 4, 5, 7, 8]

B [1, 2, 3, 4, 5, 7, 8, 6]

C [5, 3, 2, 1, 4, 7, 8]

D [1, 2, 3, 4, 5, 7, 8]

💡 Hint

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

The inorder traversal visits nodes in ascending order for this BST: 1, 2, 3, 4, 5, 7, 8.

Practice

1. What is the main advantage of Morris Traversal for inorder tree traversal compared to traditional methods?

easy

A. It only works on binary search trees

B. It is faster than all other traversal methods

C. It requires a queue to store nodes

D. It uses no extra space like stack or recursion

Solution

1. Step 1: Understand Morris Traversal space usage

   Morris Traversal modifies tree links temporarily to avoid using stack or recursion, so it uses O(1) extra space.

2. Step 2: Compare with traditional methods

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

3. Final Answer:

   It uses no extra space like stack or recursion -> Option D

4. Quick Check:

   Morris Traversal = O(1) space [OK]

Hint: Morris Traversal avoids stack and recursion for space savings [OK]

Common Mistakes:

• Thinking Morris Traversal is faster in time complexity
• Assuming it only works on binary search trees
• Believing it uses extra data structures like queues

2. Which of the following is the correct way to find the inorder predecessor of a node curr in Morris Traversal?

easy

A. Move to the left child of curr and then keep moving right until right child is null or curr

B. Move to the right child of curr until null

C. Move to the left child of curr only once

D. Move to the right child of curr and then keep moving left until left child is null

Solution

1. Step 1: Identify inorder predecessor in Morris Traversal

   The inorder predecessor is the rightmost node in the left subtree of curr.

2. Step 2: Traverse left subtree then rightmost path

   Start at curr.left, move right repeatedly until right child is null or points back to curr.

3. Final Answer:

   Move to the left child of curr and then keep moving right until right child is null or curr -> Option A

4. Quick Check:

   Inorder predecessor = left subtree's rightmost node [OK]

Hint: Inorder predecessor = left child then rightmost path [OK]

Common Mistakes:

• Moving right child first instead of left
• Stopping after one left move without going right
• Confusing predecessor with successor

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

    2
   / \
  1   3

Assuming the traversal prints node values in order.

medium

A. [1, 3, 2]

B. [2, 1, 3]

C. [1, 2, 3]

D. [3, 2, 1]

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits left subtree, then current node, then right subtree.

2. Step 2: Apply Morris traversal on given tree

   Left child is 1, root is 2, right child is 3, so output is [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option C

4. Quick Check:

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

Hint: Inorder prints left, root, right nodes [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Mixing node visit order
• Assuming traversal prints in reverse

4. Consider this snippet of Morris Traversal code:

while (curr !== null) {
  if (curr.left === null) {
    console.log(curr.val);
    curr = curr.right;
  } else {
    let pred = curr.left;
    while (pred.right !== null && pred.right !== curr) {
      pred = pred.right;
    }
    if (pred.right === null) {
      pred.right = curr;
      curr = curr.left;
    } else {
      pred.right = null;
      console.log(curr.val);
      curr = curr.right;
    }
  }
}

What is the error in this code?

medium

A. It misses printing the current node after restoring links

B. It never prints the current node's value

C. It creates infinite loop by not updating curr

D. It incorrectly sets pred.right to curr.left

Solution

1. Step 1: Check when node values are printed

   Values are printed only when curr.left === null, but after restoring links, printing is missing.

2. Step 2: Identify missing print after restoring links

   After setting pred.right = null, the code moves curr = curr.right without printing curr.val.

3. Final Answer:

   It misses printing the current node after restoring links -> Option A

4. Quick Check:

   Print missing after restoring links = bug [OK]

Hint: Print node after restoring links to avoid missing values [OK]

Common Mistakes:

• Not printing node after restoring predecessor link
• Confusing when to update curr pointer
• Incorrectly modifying tree structure

5. You want to modify Morris Traversal inorder to collect node values in an array instead of printing. Which change is necessary to correctly collect all nodes in order?

hard

A. Push curr.val only when curr.left === null

B. Push curr.val both when curr.left === null and after restoring links

C. Push curr.val only after restoring links

D. Push curr.val before moving to left child

Solution

1. Step 1: Identify when nodes are visited in Morris Traversal

   Nodes are visited either when curr.left === null or after restoring the threaded link.

2. Step 2: Collect values at both visit points

   To collect all nodes in inorder, push curr.val both when curr.left === null and after restoring links (before moving right).

3. Final Answer:

   Push curr.val both when curr.left === null and after restoring links -> Option B

4. Quick Check:

   Collect values at both visits = correct inorder [OK]

Hint: Collect node values at both visits for full inorder [OK]

Common Mistakes:

• Collecting values only once misses nodes
• Collecting before moving left causes wrong order
• Ignoring second visit after restoring links