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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to call the postorder traversal function recursively on the left child.

DSA Rust

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

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

Aright

Bparent

Cleft

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Calling postorder on the right child first.

Trying to access a non-existent field like parent.

In postorder traversal, we first visit the left child, so we call postorder on n.left.

2fill in blank

medium

Complete the code to call the postorder traversal function recursively on the right child.

DSA Rust

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

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

Aparent

Broot

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Calling postorder on left child twice.

Using incorrect field names.

After visiting the left child, postorder visits the right child, so call postorder on n.right.

3fill in blank

hard

Fix the error in printing the node value in postorder traversal.

DSA Rust

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

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

A"{}"

B"Node value:"

C"val"

Dn.val

Attempts:

3 left

💡 Hint

Common Mistakes

Printing n.val directly without a format string.

Using a string literal without {} placeholder.

The println! macro needs a format string with {} to print the value of n.val.

4fill in blank

hard

Fill both blanks to create a postorder traversal that collects values into a vector.

DSA Rust

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

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

Aleft

Bright

Cparent

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children.

Using invalid field names.

Postorder visits left child first, then right child, so blanks are filled with left and right respectively.

5fill in blank

hard

Fill all three blanks to create a postorder traversal that returns a vector of values.

DSA Rust

fn postorder_return(node: &Option<Box<TreeNode>>) -> Vec<i32> {
    if let Some(n) = node {
        let mut vals = postorder_return(&n.[1]);
        vals.extend(postorder_return(&n.[2]));
        vals.push(n.[3]);
        vals
    } else {
        Vec::new()
    }
}

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

Aleft

Bright

Cval

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using root instead of val for node value.

Swapping left and right children.

We collect values from left and right children, then push the current node's val.

Practice

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

easy

A. Left subtree, Right subtree, Root node

B. Root node, Left subtree, Right subtree

C. Root node, Right subtree, Left subtree

D. Right subtree, Left subtree, Root node

Solution

1. Step 1: Understand postorder traversal definition

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

2. Step 2: Match the order with options

   Left subtree, Right subtree, Root node matches this order exactly: Left subtree, Right subtree, Root node.

3. Final Answer:

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

4. Quick Check:

   Postorder = Left, Right, Root [OK]

Hint: Remember postorder ends with root after children [OK]

Common Mistakes:

• Confusing postorder with preorder or inorder
• Visiting root before children
• Swapping left and right subtree order

2. Which of the following Rust function signatures correctly represents a recursive postorder traversal function for a binary tree node defined as struct Node { left: Option, right: Option, value: i32 }?

easy

A. fn postorder(node: &Option) { /* visit left, right, root */ }

B. fn postorder(node: Node) { /* visit root, left, right */ }

C. fn postorder(node: &Node) -> i32 { /* visit root only */ }

D. fn postorder(node: &mut Node) -> bool { /* visit right, left, root */ }

Solution

1. Step 1: Identify correct parameter type for recursion

   Postorder traversal usually takes a reference to an optional boxed node to handle empty children safely.

2. Step 2: Match function signature with traversal order and parameter

   fn postorder(node: &Option) { /* visit left, right, root */ } uses &Option which is correct for recursive calls and mentions visiting left, right, root.

3. Final Answer:

   fn postorder(node: &Option) { /* visit left, right, root */ } -> Option A

4. Quick Check:

   Use &Option for safe recursion [OK]

Hint: Use &Option for recursive tree traversal [OK]

Common Mistakes:

• Passing ownership instead of reference
• Returning incorrect types
• Wrong traversal order in comments

3. Given the following Rust code for postorder traversal, what will be printed?

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

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

fn main() {
    let tree = Some(Box::new(Node {
        value: 1,
        left: Some(Box::new(Node {
            value: 2,
            left: None,
            right: None,
        })),
        right: Some(Box::new(Node {
            value: 3,
            left: None,
            right: None,
        })),
    }));
    postorder(&tree);
}

medium

A. 1 3 2

B. 1 2 3

C. 3 2 1

D. 2 3 1

Solution

1. Step 1: Trace postorder traversal on given tree

   The tree root is 1, left child 2, right child 3. Postorder visits left (2), right (3), then root (1).

2. Step 2: Confirm printed order

   Values printed in order: 2, then 3, then 1, matching 2 3 1 .

3. Final Answer:

   2 3 1 -> Option D

4. Quick Check:

   Postorder prints left, right, root = 2 3 1 [OK]

Hint: Print order matches left-right-root traversal [OK]

Common Mistakes:

• Confusing preorder or inorder output
• Missing spaces in print output
• Ignoring None children

4. The following Rust code attempts a postorder traversal but has a bug. What is the error?

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

medium

A. It causes a stack overflow due to infinite recursion

B. It prints root before children, not postorder

C. It does not handle None nodes correctly

D. It borrows node mutably but parameter is immutable

Solution

1. Step 1: Analyze print statement position

   The print happens before recursive calls to left and right children, so root is printed first.

2. Step 2: Compare with postorder definition

   Postorder requires visiting left and right children before root, so this order is incorrect.

3. Final Answer:

   It prints root before children, not postorder -> Option B

4. Quick Check:

   Print after children for postorder [OK]

Hint: Print root after children for correct postorder [OK]

Common Mistakes:

• Printing root before recursive calls
• Confusing preorder with postorder
• Ignoring None checks

5. You want to collect all node values of a binary tree in postorder traversal into a vector in Rust. Which of the following code snippets correctly implements this?

hard

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

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

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

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

Solution

1. Step 1: Understand postorder collection logic

   Postorder means collect left subtree values, then right subtree values, then root value.

2. Step 2: Check each code snippet for correct order and return type

   fn postorder_collect(node: &Option) -> Vec<i32> { let mut vals = Vec::new(); if let Some(n) = node { vals.extend(postorder_collect(&n.left)); vals.extend(postorder_collect(&n.right)); vals.push(n.value); } vals } correctly collects left and right subtree vectors first, then pushes root value, returning the combined vector.

3. Final Answer:

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

4. Quick Check:

   Collect left, right, then root in vector [OK]

Hint: Collect children first, then push root value [OK]

Common Mistakes:

• Pushing root before children
• Not returning combined vector
• Using mutable vector incorrectly