Morris Traversal Inorder Without Stack in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start Morris traversal by setting current to root.

DSA Javascript

let current = [1];

Drag options to blanks, or click blank then click option'

Aroot

Bnull

Chead

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Setting current to null or head instead of root.

We start Morris traversal from the root node, so current should be set to root.

2fill in blank

medium

Complete the condition to check if current has a left child.

DSA Javascript

while (current !== null) {
  if (current.[1] === null) {
    // Process current node
  }
}

Drag options to blanks, or click blank then click option'

Aright

Bchild

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking current.right instead of current.left.

We check if current.left is null to decide if we can process current or need to find predecessor.

3fill in blank

hard

Fix the error in finding the inorder predecessor by completing the loop condition.

DSA Javascript

let predecessor = current.left;
while (predecessor.[1] !== null && predecessor.[2] !== current) {
  predecessor = predecessor.right;
}

Drag options to blanks, or click blank then click option'

Achild

Bleft

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using predecessor.left instead of predecessor.right.

The inorder predecessor is found by moving to the rightmost node in the left subtree, so we check predecessor.right.

4fill in blank

hard

Fill both blanks to create or remove the temporary thread in Morris traversal.

DSA Javascript

if (predecessor.[1] === null) {
  predecessor.[2] = current;
  current = current.left;
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or parent instead of right for the thread pointer.

We create a temporary thread by setting predecessor.right to current, so both blanks are 'right'.

5fill in blank

hard

Fill all three blanks to correctly remove the temporary thread and move current in Morris traversal.

DSA Javascript

else {
  predecessor.[1] = null;
  console.log(current.[2]);
  current = current.[3];
}

Drag options to blanks, or click blank then click option'

Aright

Bvalue

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Removing thread from left pointer or printing current.left.

We remove the thread by setting predecessor.right to null, print current.value, and move current to current.right.

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