Morris Traversal Inorder Without Stack in DSA Javascript - Time & Space Complexity

Time Complexity: Morris Traversal Inorder Without Stack

O(n)

Understanding Time Complexity

We want to understand how long Morris Traversal takes to visit all nodes in a binary tree without using extra memory for a stack.

How does the number of steps grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the Morris Inorder Traversal code below.

function morrisInorder(root) {
  let current = root;
  while (current !== null) {
    if (current.left === null) {
      console.log(current.val);
      current = current.right;
    } else {
      let predecessor = current.left;
      while (predecessor.right !== null && predecessor.right !== current) {
        predecessor = predecessor.right;
      }
      if (predecessor.right === null) {
        predecessor.right = current;
        current = current.left;
      } else {
        predecessor.right = null;
        console.log(current.val);
        current = current.right;
      }
    }
  }
}
    

This code visits all nodes in a binary tree in order without using a stack or recursion by temporarily modifying the tree.

Identify Repeating Operations

Look at the loops and repeated steps:

• Primary operation: Moving through nodes and finding the rightmost child (predecessor) in the left subtree.
• How many times: Each edge in the tree is visited at most twice: once to create a temporary link and once to remove it.

How Execution Grows With Input

As the number of nodes (n) grows, the traversal visits each node and its edges a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 steps (each edge twice)
100	About 200 steps
1000	About 2000 steps

Pattern observation: The steps grow roughly twice the number of edges, so it grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to traverse grows directly in proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because of the nested while loop, the time complexity must be quadratic O(n²)."

[OK] Correct: Each edge is visited at most twice, so the total work is still proportional to the number of nodes, not squared.

Interview Connect

Understanding Morris Traversal shows you can save memory by clever tree manipulation while keeping time efficient, a skill valued in real-world coding.

Self-Check

What if we used recursion instead of Morris Traversal? How would the time and space complexity change?

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