Serialize and Deserialize Binary Tree in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new binary tree node with a given value.

DSA Rust

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

impl TreeNode {
    fn new(val: i32) -> Self {
        TreeNode {
            val: [1],
            left: None,
            right: None,
        }
    }
}

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

Aval + 1

B-1

Cval

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using a fixed value like 0 or -1 instead of the input parameter.

Modifying the input value unnecessarily.

The new node's value should be set to the input parameter 'val' to correctly initialize the node.

2fill in blank

medium

Complete the code to serialize a binary tree node's value or use '#' for None.

DSA Rust

fn serialize(root: &Option<Box<TreeNode>>, result: &mut Vec<String>) {
    match root {
        Some(node) => {
            result.push(node.val.to_string());
            serialize(&node.left, result);
            serialize(&node.right, result);
        }
        None => {
            result.push([1].to_string());
        }
    }
}

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

A"#"

B"None"

C"null"

D"0"

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'null' or 'None' as strings which are longer than needed.

Using '0' which could be confused with a valid node value.

The '#' symbol is commonly used to represent None (null) nodes in serialization.

3fill in blank

hard

Fix the error in the deserialize function to correctly parse the next token.

DSA Rust

fn deserialize(data: &mut std::slice::Iter<String>) -> Option<Box<TreeNode>> {
    if let Some(val) = data.next() {
        if val == [1] {
            return None;
        }
        let node_val = val.parse::<i32>().unwrap();
        let left = deserialize(data);
        let right = deserialize(data);
        Some(Box::new(TreeNode { val: node_val, left, right }))
    } else {
        None
    }
}

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

A"None"

B"null"

C"0"

D"#"

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 'null' or 'None' which do not match the serialization symbol.

Using '0' which is a valid node value, not a None marker.

The deserialize function must check for '#' to identify None nodes as per the serialization format.

4fill in blank

hard

Fill both blanks to complete the serialize function that returns a joined string.

DSA Rust

fn serialize_tree(root: &Option<Box<TreeNode>>) -> String {
    let mut result = Vec::new();
    serialize(root, &mut result);
    result[1]("[2]")
}

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

Ajoin

Bconcat

Cjoin_with

D""

Attempts:

3 left

💡 Hint

Common Mistakes

Using methods that do not exist on Vec like 'concat' or 'connect'.

Using a separator other than an empty string.

The Vec has a 'join' method to combine elements with a separator string, here an empty string "".

5fill in blank

hard

Fill all three blanks to complete the deserialize function that splits input and calls helper.

DSA Rust

fn deserialize_tree(data: String) -> Option<Box<TreeNode>> {
    let mut iter = data[1]("[2]")[3]();
    deserialize(&mut iter)
}

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

Asplit

Bchars

Ccollect

D""

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'split' instead of 'chars' to separate tokens.

Not collecting the iterator into a vector before passing.

We split the string into characters with 'chars', then collect into a vector for iteration.

Practice

1. What is the main purpose of serialize and deserialize functions for a binary tree?

easy

A. Sort the binary tree nodes in ascending order

B. Delete all nodes from the binary tree

C. Convert a binary tree to a string and back to a tree

D. Count the number of leaf nodes in the binary tree

Solution

1. Step 1: Understand serialization

   Serialization means converting a data structure like a binary tree into a string format that can be stored or transmitted.

2. Step 2: Understand deserialization

   Deserialization means converting the string back into the original binary tree structure.

3. Final Answer:

   Convert a binary tree to a string and back to a tree -> Option C

4. Quick Check:

   Serialize/deserialize = convert tree to string and back [OK]

Hint: Serialization means tree to string, deserialization reverses it [OK]

Common Mistakes:

• Confusing serialization with sorting
• Thinking serialization deletes nodes
• Mixing serialization with counting nodes

2. Which of the following Rust function signatures correctly represents a serialize function for a binary tree node defined as Option?

easy

A. fn serialize(root: Option) -> String

B. fn serialize(root: &TreeNode) -> Vec<i32>

C. fn serialize(root: Box<TreeNode>) -> Option<String>

D. fn serialize(root: Option<TreeNode>) -> String

Solution

1. Step 1: Check parameter type

   The binary tree root is usually represented as Option in Rust to allow empty trees and ownership.

2. Step 2: Check return type

   Serialization returns a String representing the tree.

3. Final Answer:

   fn serialize(root: Option) -> String -> Option A

4. Quick Check:

   Serialize takes Option and returns String [OK]

Hint: Serialize takes Option, returns String [OK]

Common Mistakes:

• Using &TreeNode instead of Option
• Returning Vec<i32> instead of String
• Using Option<TreeNode> without Box

3. Given the following Rust code snippet for serialization using preorder traversal, what is the output string for the tree with root value 1, left child 2, and right child 3?

fn serialize(root: Option) -> String {
    fn helper(node: &Option, s: &mut String) {
        match node {
            Some(n) => {
                s.push_str(&n.val.to_string());
                s.push(',');
                helper(&n.left, s);
                helper(&n.right, s);
            }
            None => s.push_str("#,")
        }
    }
    let mut result = String::new();
    helper(&root, &mut result);
    result
}

medium

A. "1,2,3,#,#,#,#,"

B. "1,2,#,#,3,#,#,"

C. "1,#,2,#,3,#,#,"

D. "#,1,2,3,#,#,#,"

Solution

1. Step 1: Understand preorder traversal serialization

   The function visits root, then left subtree, then right subtree, adding values and '#' for nulls.

2. Step 2: Trace the tree nodes

   Root 1: add "1," Left child 2: add "2," Left of 2 is None: add "#," Right of 2 is None: add "#," Right child 3: add "3," Left of 3 is None: add "#," Right of 3 is None: add "#,"

3. Final Answer:

   "1,2,#,#,3,#,#," -> Option B

4. Quick Check:

   Preorder serialize = "1,2,#,#,3,#,#," [OK]

Hint: Preorder serialize: root,left,right with '#' for nulls [OK]

Common Mistakes:

• Forgetting to add '#' for null children
• Mixing preorder with inorder or postorder
• Missing commas between values

4. The following Rust deserialize function has a bug. What is the main issue?

fn deserialize(data: &str) -> Option {
    fn helper(vals: &mut std::str::Split<char>) -> Option {
        match vals.next() {
            Some("#") => None,
            Some(val) => {
                let node_val = val.parse().unwrap();
                Some(Box::new(TreeNode {
                    val: node_val,
                    left: helper(vals),
                    right: helper(vals),
                }))
            }
            None => None,
        }
    }
    helper(&mut data.split(","))
}

medium

A. The function does not consume all input values

B. The function does not handle None values correctly

C. The parse unwrap can panic if val is not a number or '#'

D. The split iterator includes an empty string at the end causing parse errors

Solution

1. Step 1: Analyze split behavior

   Splitting a string like "1,2,#,#,3,#,#," by ',' produces an extra empty string at the end.

2. Step 2: Understand impact on parsing

   The helper tries to parse this empty string as a number, causing a panic or error.

3. Final Answer:

   The split iterator includes an empty string at the end causing parse errors -> Option D

4. Quick Check:

   Trailing comma causes empty string in split [OK]

Hint: Trailing comma in split creates empty string; handle or trim [OK]

Common Mistakes:

• Ignoring trailing empty string from split
• Assuming parse unwrap is always safe
• Not checking for None before parsing

5. You want to serialize a binary tree and then deserialize it back. Which approach below ensures the deserialized tree is structurally identical to the original, even if node values can be negative or zero?

hard

A. Use preorder traversal with '#' for nulls and commas as separators

B. Use inorder traversal without null markers and space separators

C. Use postorder traversal with zero as null marker and no separators

D. Use level order traversal but skip null nodes in serialization

Solution

1. Step 1: Consider traversal and null markers

   Preorder traversal with explicit '#' markers for nulls preserves tree structure exactly.

2. Step 2: Consider node values and separators

   Using commas as separators avoids confusion with negative or zero values; zero cannot be used as null marker.

3. Step 3: Evaluate other options

   Inorder without nulls loses structure; postorder with zero marker confuses zero values; skipping nulls in level order loses position info.

4. Final Answer:

   Use preorder traversal with '#' for nulls and commas as separators -> Option A

5. Quick Check:

   Preorder + '#' + commas = exact tree [OK]

Hint: Preorder + '#' + commas preserves structure with any values [OK]

Common Mistakes:

• Using inorder without null markers loses structure
• Using zero as null marker confuses zero values
• Skipping nulls in level order loses node positions