Tree Traversal Inorder Left Root Right in DSA Rust - Time & Space Complexity

Time Complexity: Tree Traversal Inorder Left Root Right

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

How does the number of steps change when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

fn inorder_traversal(node: &Option>) {
    if let Some(n) = node {
        inorder_traversal(&n.left);
        println!("{}", n.val);
        inorder_traversal(&n.right);
    }
}
    

This code visits each node in a tree in the order: left child, then root, then right child.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

Each node is visited exactly once, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Because recursion is involved, the time must be exponential or very large."

[OK] Correct: Each node is visited only once, so the total steps grow linearly, not exponentially.

Interview Connect

Understanding this traversal and its time cost helps you explain how tree operations work efficiently in many real problems.

Self-Check

"What if the tree is very unbalanced, like a linked list? How would the time complexity change?"

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Left subtree, Root node, Right subtree

B. Root node, Left subtree, Right subtree

C. Right subtree, Root node, Left subtree

D. Root node, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits nodes in the order: left subtree first, then the root node, and finally the right subtree.

2. Step 2: Match the order with options

   Left subtree, Root node, Right subtree matches the left-root-right order exactly.

3. Final Answer:

   Left subtree, Root node, Right subtree -> Option A

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder means Left, Root, then Right [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Mixing up the order of root and subtrees
• Thinking inorder visits root first

2. Which Rust code snippet correctly implements the inorder traversal function signature for a binary tree node?

easy

A. fn inorder(node: Box<TreeNode>) -> Vec<i32> { /* traversal code */ }

B. fn inorder(node: &Option) { /* traversal code */ }

C. fn inorder(node: &TreeNode) -> Vec<i32> { /* traversal code */ }

D. fn inorder(node: Option<TreeNode>) { /* traversal code */ }

Solution

1. Step 1: Understand Rust ownership and references for trees

   Binary trees in Rust often use Option to allow null and heap allocation. Passing by reference (&Option) avoids ownership transfer during traversal.

2. Step 2: Check each option's signature

   fn inorder(node: &Option) { /* traversal code */ } uses a reference to Option, which is correct for traversal without ownership moves. Others either take ownership or incorrect types.

3. Final Answer:

   fn inorder(node: &Option) { /* traversal code */ } -> Option B

4. Quick Check:

   Use &Option for safe traversal [OK]

Hint: Use references to Option for traversal [OK]

Common Mistakes:

• Taking ownership of nodes causing move errors
• Using plain TreeNode instead of Box or Option
• Returning traversal result without proper ownership

3. Given the following Rust code for inorder traversal, what is the printed output?

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

fn inorder(node: &Option) {
    if let Some(n) = node {
        inorder(&n.left);
        print!("{} ", n.val);
        inorder(&n.right);
    }
}

fn main() {
    let tree = Some(Box::new(TreeNode {
        val: 2,
        left: Some(Box::new(TreeNode {
            val: 1,
            left: None,
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: None,
        })),
    }));
    inorder(&tree);
}

medium

A. 3 2 1

B. 2 1 3

C. 1 2 3

D. 1 3 2

Solution

1. Step 1: Understand the tree structure

   The root node has value 2, left child 1, right child 3.

2. Step 2: Trace inorder traversal

   Inorder visits left subtree (1), then root (2), then right subtree (3). So output is "1 2 3 "

3. Final Answer:

   1 2 3 -> Option C

4. Quick Check:

   Inorder prints left-root-right: 1 2 3 [OK]

Hint: Inorder prints left child, root, then right child [OK]

Common Mistakes:

• Printing root before left subtree
• Confusing preorder or postorder output
• Ignoring spaces in output

4. Identify the error in this Rust inorder traversal code and select the fix:

fn inorder(node: Option) {
    if let Some(n) = node {
        inorder(n.left);
        print!("{} ", n.val);
        inorder(n.right);
    }
}

medium

A. No error, code is correct

B. Change parameter to Option<&Box<TreeNode>> and call inorder(n.left) and inorder(n.right)

C. Add return type Vec<i32> and collect values instead of printing

D. Change parameter to &Option and call inorder(&n.left) and inorder(&n.right)

Solution

1. Step 1: Identify ownership problem in parameter

   The function takes ownership of node, so recursive calls move ownership of n.left and n.right, which is invalid after first call.

2. Step 2: Fix by using references

   Changing parameter to &Option and calling inorder(&n.left) and inorder(&n.right) avoids ownership moves and allows safe recursion.

3. Final Answer:

   Change parameter to &Option and call inorder(&n.left) and inorder(&n.right) -> Option D

4. Quick Check:

   Use references to avoid ownership moves in recursion [OK]

Hint: Use references in recursive tree traversal to avoid ownership errors [OK]

Common Mistakes:

• Passing ownership causing move errors
• Trying to use Option<&Box<TreeNode>> incorrectly
• Ignoring Rust's ownership rules in recursion

5. You want to collect all values of a binary tree in inorder traversal into a vector in Rust. Which code snippet correctly implements this?

hard

A. fn inorder_collect(node: &Option) -> Vec<i32> {
let mut vals = Vec::new();
if let Some(n) = node {
vals.extend(inorder_collect(&n.left));
vals.push(n.val);
vals.extend(inorder_collect(&n.right));
}
vals
}

B. fn inorder_collect(node: Option) -> Vec<i32> {
let mut vals = Vec::new();
if let Some(n) = node {
vals.push(n.val);
vals.extend(inorder_collect(n.left));
vals.extend(inorder_collect(n.right));
}
vals
}

C. fn inorder_collect(node: &Option, vals: &mut Vec<i32>) {
if let Some(n) = node {
vals.push(n.val);
inorder_collect(&n.left, vals);
inorder_collect(&n.right, vals);
}
}

D. fn inorder_collect(node: &Option) -> Vec<i32> {
let mut vals = Vec::new();
if let Some(n) = node {
vals.push(n.val);
vals.extend(inorder_collect(&n.right));
vals.extend(inorder_collect(&n.left));
}
vals
}

Solution

1. Step 1: Understand inorder traversal collection

   Inorder means collect left subtree values first, then root value, then right subtree values.

2. Step 2: Check each code snippet's order and ownership

   fn inorder_collect(node: &Option) -> Vec<i32> { let mut vals = Vec::new(); if let Some(n) = node { vals.extend(inorder_collect(&n.left)); vals.push(n.val); vals.extend(inorder_collect(&n.right)); } vals } correctly uses references, collects left subtree, then root, then right subtree, returning combined vector.

3. Final Answer:

   fn inorder_collect(node: &Option) -> Vec<i32> { let mut vals = Vec::new(); if let Some(n) = node { vals.extend(inorder_collect(&n.left)); vals.push(n.val); vals.extend(inorder_collect(&n.right)); } vals } -> Option A

4. Quick Check:

   Collect left, root, right in vector [OK]

Hint: Collect left subtree, then root, then right subtree values [OK]

Common Mistakes:

• Wrong order of pushing values
• Taking ownership instead of references
• Appending right subtree before left