Tree Traversal Postorder Left Right Root in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of postorder traversal?

Given the binary tree below, what is the output of the postorder traversal (Left, Right, Root)? Tree structure: 1 / \ 2 3 / \ 4 5

DSA Rust

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

fn postorder(root: &Option<Box<Node>>, result: &mut Vec<i32>) {
    if let Some(node) = root {
        postorder(&node.left, result);
        postorder(&node.right, result);
        result.push(node.val);
    }
}

fn main() {
    let root = Some(Box::new(Node {
        val: 1,
        left: Some(Box::new(Node {
            val: 2,
            left: Some(Box::new(Node { val: 4, left: None, right: None })),
            right: Some(Box::new(Node { val: 5, left: None, right: None })),
        })),
        right: Some(Box::new(Node { val: 3, left: None, right: None })),
    }));

    let mut result = Vec::new();
    postorder(&root, &mut result);
    println!("{:?}", result);
}

A [4, 5, 2, 3, 1]

B [1, 2, 3, 4, 5]

C [1, 3, 2, 4, 5]

D [2, 4, 5, 3, 1]

💡 Hint

Remember postorder means visit left subtree, then right subtree, then the root node.

Postorder traversal visits nodes in Left, Right, Root order. For the given tree, left subtree of root 1 is subtree rooted at 2, which itself has left 4 and right 5. So traversal order is 4, 5, 2, then right subtree 3, then root 1.

🧠 Conceptual

intermediate

Which traversal order matches postorder?

Select the correct order of visiting nodes in postorder traversal of a binary tree.

A Right, Left, Root

B Root, Left, Right

C Left, Root, Right

D Left, Right, Root

💡 Hint

Postorder traversal visits children before the parent.

Postorder traversal visits the left subtree first, then the right subtree, and finally the root node.

🔧 Debug

advanced

Identify the error in this postorder traversal code

What error will this Rust code produce when compiling? Code snippet: fn postorder(root: &Option>, result: &mut Vec) { if root.is_some() { postorder(&root.as_ref().unwrap().left, result); postorder(&root.as_ref().unwrap().right, result); result.push(root.unwrap().val); } }

DSA Rust

fn postorder(root: &Option<Box<Node>>, result: &mut Vec<i32>) {
    if root.is_some() {
        postorder(&root.as_ref().unwrap().left, result);
        postorder(&root.as_ref().unwrap().right, result);
        result.push(root.unwrap().val);
    }
}

A Compile error: mismatched types

B Compile error: cannot move out of borrowed content

C No error, runs correctly

D Runtime panic: unwrap on None value

💡 Hint

Check how unwrap() is used on a borrowed reference.

Calling unwrap() on &Option> tries to move the value out of a borrowed reference, which Rust forbids. This causes a compile error.

🚀 Application

advanced

How many nodes are visited in postorder traversal?

Given a binary tree with 7 nodes, how many nodes will be visited during a postorder traversal?

A 6

B 8

C 7

D Depends on tree shape

💡 Hint

Traversal visits every node exactly once.

Postorder traversal visits every node in the tree exactly once, so the count equals the total nodes.

❓ Predict Output

expert

What is the output of this postorder traversal with missing children?

Consider the binary tree below with some missing children: 10 / 5 \ 7 What is the output of the postorder traversal (Left, Right, Root)?

DSA Rust

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

fn postorder(root: &Option<Box<Node>>, result: &mut Vec<i32>) {
    if let Some(node) = root {
        postorder(&node.left, result);
        postorder(&node.right, result);
        result.push(node.val);
    }
}

fn main() {
    let root = Some(Box::new(Node {
        val: 10,
        left: Some(Box::new(Node {
            val: 5,
            left: None,
            right: Some(Box::new(Node { val: 7, left: None, right: None })),
        })),
        right: None,
    }));

    let mut result = Vec::new();
    postorder(&root, &mut result);
    println!("{:?}", result);
}

A [7, 5, 10]

B [5, 7, 10]

C [7, 10, 5]

D [10, 5, 7]

💡 Hint

Remember to visit left subtree, then right subtree, then root.

Node 5 has no left child but has right child 7. Postorder visits left (none), then right (7), then node 5, then root 10.

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