Why Morris Traversal Inorder Without Stack in DSA Rust?

The Big Idea

Discover how to walk through a tree without carrying any extra memory--like magic!

The Scenario

Imagine you want to read all the pages of a book, but you can only hold one page at a time and have no bookmark to remember where you left off.

You try to flip back and forth manually, but it's confusing and slow.

The Problem

Using a manual method like a stack to remember pages means extra memory and effort.

It's like carrying a pile of bookmarks, which can get lost or mixed up.

This slows you down and makes the process error-prone.

The Solution

Morris Traversal cleverly uses the book's own pages to mark where to return next, without extra bookmarks.

It temporarily changes the book's structure to remember the path, then restores it perfectly.

This way, you can read all pages in order without extra memory or confusion.

Before vs After

✗ Before

unsafe fn inorder_traversal(root: *mut TreeNode) {
    let mut stack: Vec<*mut TreeNode> = Vec::new();
    let mut current = root;
    while !current.is_null() || !stack.is_empty() {
        while !current.is_null() {
            stack.push(current);
            current = (*current).left;
        }
        current = *stack.last().unwrap_or(&std::ptr::null_mut());
        stack.pop();
        println!("{}", (*current).val);
        current = (*current).right;
    }
}

✓ After

unsafe fn morris_inorder_traversal(root: *mut TreeNode) {
    let mut current = root;
    while !current.is_null() {
        if (*current).left.is_null() {
            println!("{}", (*current).val);
            current = (*current).right;
        } else {
            let mut predecessor = (*current).left;
            while !(*predecessor).right.is_null() && (*(*predecessor).right).val != (*current).val {
                predecessor = (*predecessor).right;
            }
            if (*predecessor).right.is_null() {
                (*predecessor).right = current;
                current = (*current).left;
            } else {
                (*predecessor).right = std::ptr::null_mut();
                println!("{}", (*current).val);
                current = (*current).right;
            }
        }
    }
}

What It Enables

This method enables traversing a tree in order using no extra memory, making it efficient and elegant.

Real Life Example

Think of exploring a maze where you mark your path by temporarily placing stones, then remove them as you backtrack, so you never get lost and don't carry extra tools.

Key Takeaways

Morris Traversal avoids extra memory by temporarily modifying the tree.

It restores the tree after traversal, keeping data intact.

This method is efficient for memory-limited environments.

Practice

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

easy

A. It is faster than all other traversal methods

B. It requires a special tree structure

C. It only works on binary search trees

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.

2. Step 2: Compare with traditional methods

   Traditional inorder traversal uses stack or recursion which require extra space. Morris Traversal uses O(1) extra space.

3. Final Answer:

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

4. Quick Check:

   Space complexity = O(1) [OK]

Hint: Remember Morris uses threaded links, no stack needed [OK]

Common Mistakes:

• Thinking Morris is faster in time complexity
• Assuming it only works on BSTs
• Confusing it with recursive traversal

2. Which Rust code snippet correctly initializes a mutable pointer to the root node for Morris Traversal?

easy

A. let mut current = root.as_ref();

B. let mut current = root.as_mut();

C. let current = root.clone();

D. let current = &root;

Solution

1. Step 1: Identify mutable pointer requirement

   Morris Traversal modifies tree links, so mutable access is needed.

2. Step 2: Choose correct Rust method

   root.as_mut() gives mutable reference, root.as_ref() is immutable, clone copies, &root is immutable reference.

3. Final Answer:

   let mut current = root.as_mut(); -> Option B

4. Quick Check:

   Mutable reference needed = let mut current = root.as_mut(); [OK]

Hint: Use as_mut() for mutable access in Rust trees [OK]

Common Mistakes:

• Using as_ref() which is immutable
• Cloning unnecessarily
• Using immutable references when mutation is needed

3. Given the following Rust code for Morris Traversal on a binary tree with nodes 1 (root), 2 (left child), and 3 (right child), what is the printed inorder sequence?

struct Node {
    val: i32,
    left: Option,
    right: Option,
}

fn morris_inorder(root: &mut Option) {
    let mut current = root.as_mut();
    while let Some(node) = current {
        if node.left.is_none() {
            print!("{} ", node.val);
            current = node.right.as_mut();
        } else {
            let mut pre = node.left.as_mut();
            while let Some(p) = pre {
                if p.right.is_some() && p.right.as_ref().unwrap().val != node.val {
                    pre = p.right.as_mut();
                } else {
                    break;
                }
            }
            if pre.is_some() && pre.as_ref().unwrap().right.is_none() {
                pre.as_mut().unwrap().right = Some(Box::new(Node { val: node.val, left: None, right: None }));
                current = node.left.as_mut();
            } else {
                pre.as_mut().unwrap().right = None;
                print!("{} ", node.val);
                current = node.right.as_mut();
            }
        }
    }
}

medium

A. 2 1 3

B. 1 2 3

C. 3 1 2

D. 1 3 2

Solution

1. Step 1: Understand tree structure

   Root=1, left=2, right=3. Inorder is left-root-right.

2. Step 2: Trace Morris traversal output

   Visit left (2), then root (1), then right (3). So output is "2 1 3".

3. Final Answer:

   2 1 3 -> Option A

4. Quick Check:

   Inorder traversal = 2 1 3 [OK]

Hint: Inorder = left, root, right; Morris prints in this order [OK]

Common Mistakes:

• Confusing preorder or postorder output
• Not restoring links causing infinite loops
• Misreading tree structure

4. In the following Rust snippet for Morris Traversal, what is the main bug causing an infinite loop?

fn morris_inorder_bug(root: &mut Option) {
    let mut current = root.as_mut();
    while let Some(node) = current {
        if node.left.is_some() {
            let mut pre = node.left.as_mut();
            while pre.is_some() && pre.as_ref().unwrap().right.is_some() {
                pre = pre.unwrap().right.as_mut();
            }
            if pre.is_some() {
                pre.as_mut().unwrap().right = Some(Box::new(Node { val: node.val, left: None, right: None }));
                current = node.left.as_mut();
            }
        } else {
            print!("{} ", node.val);
            current = node.right.as_mut();
        }
    }
}

medium

A. Using immutable references instead of mutable

B. Incorrectly printing node values

C. Not removing temporary threaded links after use

D. Missing base case for empty tree

Solution

1. Step 1: Analyze threaded link handling

   Code creates temporary right links but never removes them.

2. Step 2: Understand infinite loop cause

   Without removing threaded links, traversal cycles endlessly on same nodes.

3. Final Answer:

   Not removing temporary threaded links after use -> Option C

4. Quick Check:

   Threaded links must be removed to avoid loops [OK]

Hint: Always remove threaded links after visiting left subtree [OK]

Common Mistakes:

• Ignoring threaded link removal
• Confusing printing with traversal logic
• Not using mutable references properly

5. You want to modify Morris Traversal inorder code in Rust to collect node values into a vector instead of printing. Which change correctly implements this without breaking the traversal logic?

fn morris_inorder_collect(root: &mut Option) -> Vec<i32> {
    let mut result = Vec::new();
    let mut current = root.as_mut();
    while let Some(node) = current {
        if node.left.is_none() {
            // Change here
            current = node.right.as_mut();
        } else {
            let mut pre = node.left.as_mut();
            while let Some(p) = pre {
                if p.right.is_some() && p.right.as_ref().unwrap().val != node.val {
                    pre = p.right.as_mut();
                } else {
                    break;
                }
            }
            if pre.is_some() && pre.as_ref().unwrap().right.is_none() {
                pre.as_mut().unwrap().right = Some(Box::new(Node { val: node.val, left: None, right: None }));
                current = node.left.as_mut();
            } else {
                pre.as_mut().unwrap().right = None;
                // Change here
                current = node.right.as_mut();
            }
        }
    }
    result
}

hard

A. Replace both comments with result.push(node.val);

B. Replace first comment with print!("{} ", node.val); second with result.push(node.val);

C. Replace first comment with result.push(node.val); second with print!("{} ", node.val);

D. Replace both comments with print!("{} ", node.val);

Solution

1. Step 1: Identify where node values are processed

   Values are printed at two places: when left is None and after restoring threaded links.

2. Step 2: Replace printing with pushing to vector

   To collect values, replace both prints with result.push(node.val) to keep traversal logic intact.

3. Final Answer:

   Replace both comments with result.push(node.val); -> Option A

4. Quick Check:

   Collect values by pushing at both print points [OK]

Hint: Push values where prints occur to collect inorder sequence [OK]

Common Mistakes:

• Mixing print and push causing partial collection
• Pushing only once missing some nodes
• Breaking traversal by adding prints