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

Recall & Review

beginner

What is inorder traversal in a binary tree?

Inorder traversal visits nodes in this order: left child, root node, then right child. It processes the left subtree first, then the root, then the right subtree.

beginner

Why is inorder traversal useful for binary search trees (BST)?

Inorder traversal of a BST visits nodes in ascending order of their values, which helps to get sorted data from the tree.

intermediate

In Rust, what is a common way to represent a binary tree node?

A binary tree node is often represented as a struct with a value and optional boxed left and right children, for example:
struct Node {
value: i32,
left: Option,
right: Option,
}

beginner

What is the base case in a recursive inorder traversal function?

The base case is when the current node is None (empty). Then the function returns without doing anything, stopping recursion.

beginner

What is the order of node visits in inorder traversal?

Left subtree nodes are visited first, then the root node, and finally the right subtree nodes.

In inorder traversal, which node is visited second?

A Left child

B Right child

C Root node

D None

In inorder traversal, the root node is visited after the left child.

What does inorder traversal of a binary search tree produce?

A Nodes in descending order

B Nodes in random order

C Only leaf nodes

D Nodes in ascending order

Inorder traversal visits nodes in ascending order for a binary search tree.

What is the first step in inorder traversal of a node?

A Visit the right child

B Visit the left child

C Visit the root node

D Return immediately

In inorder traversal, the left child is visited first.

In Rust, how do you represent a child node that might be missing?

A Using Option<Box<Node>>

B Using &Node

C Using Vec<Node>

D Using i32

Option> allows representing a child node that can be None or Some node.

What stops the recursion in an inorder traversal function?

A When node is None

B When node is root

C When node has both children

D When node is Some

Recursion stops when the current node is None (empty).

Explain how inorder traversal works on a binary tree.

Think about the order of visiting left, root, and right nodes.

Describe how you would implement inorder traversal in Rust for a binary tree node struct.

Focus on Rust's Option type and recursion.

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