Serialize and Deserialize Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start serialization by checking if the root is null.

DSA C++

string serialize(TreeNode* root) {
    if (root == [1]) return "#";
    // rest of serialization code
}

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

ANULL

Bnullptr

C0

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or NULL instead of nullptr

Checking root == root which is always true

We check if the root pointer is nullptr to know if the tree or subtree is empty.

2fill in blank

medium

Complete the code to append the root value to the serialized string.

DSA C++

string serialize(TreeNode* root) {
    if (root == nullptr) return "#";
    return to_string(root->val) + [1] + serialize(root->left) + [1] + serialize(root->right);
}

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

A";"

B" "

C","

D"-"

Attempts:

3 left

💡 Hint

Common Mistakes

Using spaces or semicolons which complicate parsing

No separator causing concatenated values

We use a comma to separate node values in the serialized string for easy splitting later.

3fill in blank

hard

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

DSA C++

string nextToken;
getline(ss, [1], ',');

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

AnextToken

Btoken

Cvalue

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undeclared variable

Using the wrong variable name causing compilation errors

The variable nextToken is used to store the next substring from the stream.

4fill in blank

hard

Fill both blanks to correctly build the tree nodes during deserialization.

DSA C++

if (nextToken == "#") return [1];
int val = stoi(nextToken);
TreeNode* node = new TreeNode(val);
node->left = deserializeHelper(ss);
node->right = [2];
return node;

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

Anullptr

BdeserializeHelper(ss)

Cnew TreeNode(val)

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Returning new nodes for null

Not calling deserializeHelper for right subtree

Return nullptr for null nodes and recursively call deserializeHelper for right subtree.

5fill in blank

hard

Fill all three blanks to complete the deserialize function using a stringstream.

DSA C++

TreeNode* deserialize(string data) {
    stringstream [1](data);
    return [2]([3]);
}

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

Ass

BdeserializeHelper

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names

Passing string instead of stringstream to helper

Create a stringstream named ss from data and call deserializeHelper(ss) to build the tree.

Practice

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

easy

A. Find the height of the binary tree

B. Sort the binary tree nodes in ascending order

C. Delete all nodes of the binary tree

D. Convert a binary tree to a string and back to a 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 binary tree -> Option D

4. Quick Check:

   Serialize/Deserialize = Convert tree to/from string [OK]

Hint: Think of serialize as packing, deserialize as unpacking [OK]

Common Mistakes:

• Confusing serialization with sorting
• Thinking it deletes nodes
• Assuming it calculates tree height

2. Which of the following is a correct way to represent a null node during serialization of a binary tree in C++ using a single-character marker?

easy

A. "null"

B. "#"

C. 0

D. -1

Solution

1. Step 1: Identify common null markers

   In many serialization methods, a special character like '#' is used to mark null nodes to distinguish them from valid node values.

2. Step 2: Compare options

   "null" is a string but longer and less common; 0 and -1 can be valid node values, so they are ambiguous.

3. Final Answer:

   "#" -> Option B

4. Quick Check:

   Null node marker = "#" [OK]

Hint: Use a unique symbol like '#' for null nodes [OK]

Common Mistakes:

• Using 0 or -1 which can be valid node values
• Using the word "null" which is verbose
• Not marking null nodes at all

3. Given the following serialized string of a binary tree using preorder traversal with '#' as null marker: "1 2 # # 3 4 # # 5 # #", what is the inorder traversal of the tree?

medium

A. 2 1 4 3 5

B. 1 2 3 4 5

C. 2 4 1 5 3

D. 3 2 1 5 4

Solution

1. Step 1: Reconstruct the tree from preorder

   Preorder: 1 (root), 2 (left child), # (left null), # (right null), 3 (right child), 4 (left child), #, #, 5 (right child), #, #.

2. Step 2: Draw the tree and find inorder

   Tree structure:
       1
     /     \
    2        3
             /  \
           4      5
   Inorder (left, root, right): 2 1 4 3 5

3. Final Answer:

   2 1 4 3 5 -> Option A

4. Quick Check:

   Inorder from tree = 2 1 4 3 5 [OK]

Hint: Draw tree from preorder, then do inorder traversal [OK]

Common Mistakes:

• Mixing preorder and inorder sequences
• Ignoring null markers
• Incorrect tree reconstruction

4. Consider this C++ snippet for deserializing a binary tree from a vector of strings representing preorder traversal with '#' as null marker:

TreeNode* deserializeHelper(vector<string>& data, int& index) {
  if (index >= data.size() || data[index] == "#") {
    index++;
    return nullptr;
  }
  TreeNode* node = new TreeNode(stoi(data[index++]));
  node->left = deserializeHelper(data, index);
  node->right = deserializeHelper(data, index);
  return node;
}

What is the bug in this code?

medium

A. The stoi conversion can throw exception without try-catch

B. The index is incremented twice for null nodes causing skipping elements

C. The function does not handle empty input vector

D. The function does not assign right child correctly

Solution

1. Step 1: Identify the stoi vulnerability

   The code uses stoi(data[index++]) without validating that data[index] is a valid integer string. Malformed input (non-numeric other than '#') causes unhandled std::invalid_argument exception.

2. Step 2: Rule out other options

   Index increments once per token: for '#' via index++ in if; for values via index++ in stoi. Empty input handled by index >= data.size(). Right child assigned correctly.

3. Final Answer:

   The stoi conversion can throw exception without try-catch -> Option A

4. Quick Check:

   stoi requires valid numeric strings [OK]

Hint: Validate inputs before stoi to prevent exceptions [OK]

Common Mistakes:

• Assuming no index increment for null nodes
• Ignoring possible exceptions from stoi
• Thinking right child assignment is missing

5. You want to serialize a binary tree and then deserialize it back. However, your tree contains integer node values that can be negative, zero, or positive. Which approach ensures correct serialization and deserialization without ambiguity?

hard

A. Serialize only node values ignoring nulls, reconstruct tree by sorting

B. Use comma-separated values but skip null markers to reduce size

C. Use space-separated values with '#' for nulls and convert strings to int using stoi

D. Serialize using preorder but replace negative values with zero

Solution

1. Step 1: Consider node values including negatives and zero

   Using space-separated values with '#' for nulls preserves structure and allows negative values as strings can represent negatives.

2. Step 2: Evaluate other options

   Use comma-separated values but skip null markers to reduce size skips nulls, losing tree shape.
   Serialize only node values ignoring nulls, reconstruct tree by sorting ignores nulls and sorts, losing original structure.
   Serialize using preorder but replace negative values with zero replaces negatives with zero, losing data.

3. Final Answer:

   Use space-separated values with '#' for nulls and convert strings to int using stoi -> Option C

4. Quick Check:

   Preserve nulls and convert strings properly [OK]

Hint: Keep null markers and convert strings carefully [OK]

Common Mistakes:

• Skipping null markers losing tree shape
• Replacing negative values losing data
• Sorting nodes losing original tree