Tree Traversal Inorder Left Root Right in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to visit the left child first in inorder traversal.

DSA Rust

fn inorder(node: &Option<Box<TreeNode>>) {
    if let Some(n) = node {
        inorder(&n.[1]);
        print!("{} ", n.val);
        inorder(&n.right);
    }
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' causes wrong traversal order.

Trying to access 'val' as a child causes errors.

In inorder traversal, we first visit the left child, so we call inorder(&n.left).

2fill in blank

medium

Complete the code to print the current node's value in inorder traversal.

DSA Rust

fn inorder(node: &Option<Box<TreeNode>>) {
    if let Some(n) = node {
        inorder(&n.left);
        print!("{} ", n.[1]);
        inorder(&n.right);
    }
}

Drag options to blanks, or click blank then click option'

Aleft

Bval

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Printing left or right instead of val causes wrong output.

Printing parent causes compilation error.

We print the current node's value using n.val in inorder traversal after visiting left child.

3fill in blank

hard

Fix the error in the recursive call to visit the right child in inorder traversal.

DSA Rust

fn inorder(node: &Option<Box<TreeNode>>) {
    if let Some(n) = node {
        inorder(&n.left);
        print!("{} ", n.val);
        inorder(&n.[1]);
    }
}

Drag options to blanks, or click blank then click option'

Aleft

Bparent

Cval

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' again causes infinite recursion.

Using 'val' or 'parent' causes errors.

To complete inorder traversal, we must visit the right child using inorder(&n.right).

4fill in blank

hard

Fill both blanks to create a recursive inorder traversal that returns a vector of values.

DSA Rust

fn inorder_collect(node: &Option<Box<TreeNode>>, vals: &mut Vec<i32>) {
    if let Some(n) = node {
        inorder_collect(&n.[1], vals);
        vals.push(n.[2]);
        inorder_collect(&n.right, vals);
    }
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right causes wrong order.

Using 'right' or 'parent' in blanks causes errors.

We recursively visit left child (n.left) and push the current node's value (n.val) to the vector.

5fill in blank

hard

Fill all three blanks to create a function that returns a vector of inorder traversal values using map and flatten.

DSA Rust

fn inorder_values(node: &Option<Box<TreeNode>>) -> Vec<i32> {
    node.as_ref().map(|n| {
        let mut left_vals = inorder_values(&n.[1]);
        left_vals.push(n.[2]);
        left_vals.extend(inorder_values(&n.[3]));
        left_vals
    }).unwrap_or_else(Vec::new)
}

Drag options to blanks, or click blank then click option'

Aleft

Bval

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up left and right causes wrong order.

Using 'parent' causes compilation errors.

We collect values from left child (n.left), then current node's value (n.val), then right child (n.right).

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