Tree Traversal Inorder Left Root Right in DSA Rust - Build from Scratch

Tree Traversal Inorder Left Root Right

📖 Scenario: Imagine you have a family tree stored in a computer program. You want to visit each family member in a special order: first the left child, then the parent, then the right child. This is called inorder traversal.

🎯 Goal: You will build a simple binary tree and write code to visit its nodes in inorder (left, root, right) order, printing each node's value.

📋 What You'll Learn

Create a binary tree node struct with integer values

Build a small binary tree with 3 nodes: root 2, left child 1, right child 3

Write a recursive function called inorder_traversal that visits nodes in left-root-right order

Print the node values in inorder traversal separated by spaces

💡 Why This Matters

🌍 Real World

Tree traversal is used in many applications like searching family trees, organizing files, and parsing expressions.

💼 Career

Understanding tree traversal is important for software engineers working with data structures, databases, and compilers.

1

Create the binary tree nodes

Create a struct called TreeNode with fields val: i32, left: Option>, and right: Option>. Then create a variable called root that is a TreeNode with value 2, left child with value 1, and right child with value 3.

DSA Rust

// Define TreeNode struct and create root node with left and right children
// Your code here

Hint

Use Option> for left and right to allow empty children.

2

Add the inorder traversal function

Add a recursive function called inorder_traversal that takes a reference to an Option> and prints the node values in inorder (left, root, right) order separated by spaces.

DSA Rust

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

// Define inorder_traversal function here
// Your code here

fn main() {
    let root = 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 })),
    };
}

Hint

Use if let Some(n) = node to check if the node exists before visiting children.

3

Call inorder traversal on the root

Inside main, call the inorder_traversal function with a reference to the root wrapped in Some(Box::new(root)).

DSA Rust

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

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

fn main() {
    let root = 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 })),
    };
    // Call inorder_traversal here
    // Your code here
}

Hint

Wrap root in Some(Box::new(...)) to match the function parameter type.

4

Print the final inorder traversal output

Add a println!() after the inorder_traversal call in main to print a newline so the output looks clean.

DSA Rust

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

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

fn main() {
    let root = 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_traversal(&Some(Box::new(root)));
    // Print newline here
    // Your code here
}

Hint

Use println!() to add a newline after printing all node values.

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