Tree Traversal Postorder Left Right Root in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the order of visiting nodes in Postorder Tree Traversal?

In Postorder traversal, nodes are visited in this order: Left subtree, Right subtree, then Root node.

intermediate

Why is Postorder traversal useful in expression trees?

Postorder traversal visits children before the parent, which matches how expressions are evaluated (operands before operators).

intermediate

Rust code snippet: What does this Postorder traversal function print? 

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

This function prints the values of the tree nodes in Postorder: left child first, then right child, then the node itself.

beginner

In Postorder traversal, when do we visit the root node?

The root node is visited last, after visiting both left and right subtrees.

beginner

What is the output of Postorder traversal on this tree? 1 / \ 2 3 / \ 4 5

The output is: 4 -> 5 -> 2 -> 3 -> 1

In Postorder traversal, which node is visited first?

A Left child

B Root node

C Right child

D None

Postorder traversal always visits the left subtree first.

What is the correct Postorder traversal sequence for a single node tree?

A No nodes visited

B Left, Right, Root

C Right, Left, Root

D Root only

With only one node, Postorder visits the root last, which is also the only node.

Which traversal visits the root node last?

A Postorder

B Preorder

C Inorder

D Level order

Postorder traversal visits root after left and right subtrees.

In Rust, which keyword helps safely access an Option value in Postorder traversal?

A unwrap()

B panic!

C if let

D match

"if let" is used to safely check and access Some value inside Option.

What is the Postorder traversal output for this tree? 10 \ 20

A 10 -> 20 -> null

B 20 -> 10

C 10 -> null

D 20 -> null

Postorder visits left (none), right (20), then root (10). So output is 20 -> 10.

Explain how Postorder traversal visits nodes in a binary tree.

Think about the order Left, Right, Root.

Describe a real-life example where Postorder traversal is useful.

Consider how tasks depend on subtasks.

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