Tree Traversal Postorder Left Right Root in DSA Rust - Time & Space Complexity

Time Complexity: Tree Traversal Postorder Left Right Root

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

Specifically, we ask: How does postorder traversal scale with the number of nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code visits each node in the tree after visiting its left and right children, printing values in postorder.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

Each node is visited once, so the total work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Postorder traversal takes more than linear time because it visits nodes multiple times."

[OK] Correct: Each node is visited exactly once, even though recursion happens, so the total work is linear.

Interview Connect

Understanding traversal time helps you explain how tree algorithms scale, a key skill for many coding challenges.

Self-Check

"What if we changed the traversal to visit nodes multiple times, like in some backtracking algorithms? How would the time complexity change?"

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