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

Time Complexity: Morris Traversal Inorder Without Stack

O(n)

Understanding Time Complexity

We want to understand how the Morris Traversal method for inorder tree traversal performs as the tree grows.

Specifically, how the number of steps changes when the tree has more nodes.

Scenario Under Consideration

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

public void morrisInorder(TreeNode root) {
    TreeNode current = root;
    while (current != null) {
        if (current.left == null) {
            System.out.print(current.val + " ");
            current = current.right;
        } else {
            TreeNode 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;
                System.out.print(current.val + " ");
                current = current.right;
            }
        }
    }
}
    

This code traverses a binary tree inorder without using extra stack or recursion by temporarily modifying pointers.

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) increases, each node is visited a limited number of times.

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

Pattern observation: The operations grow roughly linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Since there is a nested while loop, the time complexity must be O(n²)."

[OK] Correct: The inner loop moves along edges that are only visited twice total, so the total work is still proportional to n, not n squared.

Interview Connect

Understanding Morris Traversal shows you can traverse trees efficiently without extra memory, a skill that impresses by showing deep knowledge of tree algorithms.

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 the traditional recursive or stack-based methods?

easy

A. It requires the tree to be balanced

B. It is faster than all other traversal methods

C. It only works on binary search trees

D. It uses no extra space for 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 recursion or stack which require O(h) space, where h is tree height.

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Morris uses threaded links to avoid stack [OK]

Common Mistakes:

• Thinking Morris is faster in time complexity
• Assuming it only works on BSTs
• Believing it requires balanced trees

2. Which of the following code snippets correctly creates the temporary thread link in Morris Traversal inorder method?

easy

A. current.right = predecessor;

B. current.left = predecessor;

C. predecessor.right = current;

D. predecessor.left = current;

Solution

1. Step 1: Identify the thread link direction

   In Morris Traversal, the right child of the inorder predecessor is linked to current node to create a thread.

2. Step 2: Match with code snippet

   The correct assignment is predecessor.right = current; to create the temporary link.

3. Final Answer:

   predecessor.right = current; -> Option C

4. Quick Check:

   Thread link = predecessor.right = current [OK]

Hint: Thread link always sets predecessor.right to current [OK]

Common Mistakes:

• Assigning current.left to predecessor instead
• Using predecessor.left instead of right
• Confusing current.right with predecessor.right

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

    2
   / \
  1   3

medium

A. [1, 2, 3]

B. [2, 1, 3]

C. [3, 2, 1]

D. [1, 3, 2]

Solution

1. Step 1: Perform inorder traversal manually

   Inorder traversal visits left subtree, root, then right subtree. For this tree: left=1, root=2, right=3.

2. Step 2: List nodes in inorder

   Nodes visited in order: 1, 2, 3.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Confusing preorder or postorder with inorder
• Mixing up left and right child order
• Assuming Morris changes traversal order

4. Consider this buggy snippet from a Morris inorder traversal implementation:

while (current != null) {
  if (current.left == null) {
    System.out.print(current.val + " ");
    current = current.right;
  } else {
    TreeNode predecessor = current.left;
    while (predecessor.right != null) {
      predecessor = predecessor.right;
    }
    predecessor.right = current;
    current = current.left;
  }
}

What is the main bug causing an infinite loop?

medium

A. Not printing current.val when left is not null

B. Not checking if predecessor.right == current before linking

C. Not moving current to current.right when left is null

D. Not initializing predecessor correctly

Solution

1. Step 1: Identify missing condition in inner loop

   The inner while loop should stop if predecessor.right == current to avoid infinite loop.

2. Step 2: Explain infinite loop cause

   Without checking predecessor.right == current, the loop never breaks and keeps moving right, causing infinite loop.

3. Final Answer:

   Not checking if predecessor.right == current before linking -> Option B

4. Quick Check:

   Check predecessor.right == current to break loop [OK]

Hint: Always check predecessor.right == current to avoid loops [OK]

Common Mistakes:

• Ignoring thread existence check
• Assuming printing happens in else block
• Incorrect predecessor initialization

5. Given a binary tree where some nodes have only left children, how does Morris Traversal handle these nodes to produce the inorder sequence without extra space?

hard

A. It creates temporary right links from the rightmost node of left subtree back to current node

B. It skips nodes with only left children to avoid infinite loops

C. It uses recursion internally to handle left-only nodes

D. It converts the tree into a linked list before traversal

Solution

1. Step 1: Understand Morris Traversal threading

   For nodes with left children, Morris Traversal finds the rightmost node in left subtree and links its right pointer to current node temporarily.

2. Step 2: Explain handling of left-only nodes

   This threading allows traversal to return to current node after left subtree is processed, enabling inorder traversal without stack or recursion.

3. Final Answer:

   It creates temporary right links from the rightmost node of left subtree back to current node -> Option A

4. Quick Check:

   Threading rightmost left subtree node to current [OK]

Hint: Thread rightmost left child back to current node [OK]

Common Mistakes:

• Thinking recursion is used internally
• Assuming nodes are skipped
• Believing tree is converted to linked list