Serialize and Deserialize Binary Tree in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does it mean to serialize a binary tree?

Serialization means converting a binary tree into a string or list format so it can be stored or sent easily.

beginner

What is the purpose of deserialization in binary trees?

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

intermediate

Why do we use a special marker like '#' or 'null' during serialization?

We use markers to represent empty (null) child nodes so the tree structure can be fully recovered during deserialization.

intermediate

Which tree traversal method is commonly used for serialization and why?

Preorder traversal is commonly used because it records the root first, then left and right subtrees, making it easy to reconstruct the tree.

intermediate

What is the time complexity of serializing and deserializing a binary tree?

Both serialization and deserialization take O(n) time, where n is the number of nodes, because each node is visited once.

What does the '#' symbol usually represent in serialized binary trees?

A A duplicate node

B A null or empty child node

C The root node

D A leaf node

The '#' symbol is used to mark null or empty child nodes during serialization.

Which traversal order is best suited for serializing a binary tree to preserve structure?

A Preorder

B Inorder

C Postorder

D Level order

Preorder traversal records root first, then left and right children, which helps in reconstructing the tree.

What is the main goal of deserializing a binary tree?

A Convert string back to original tree structure

B Print the tree nodes

C Count the number of nodes

D Balance the tree

Deserialization rebuilds the original tree from the serialized string.

If a binary tree has 7 nodes, what is the time complexity to serialize it?

A O(n^2)

B O(log 7)

C O(1)

D O(7)

Serialization visits each node once, so time is proportional to number of nodes, O(n).

Why can't we use inorder traversal alone for serialization?

A It is slower than preorder

B It skips leaf nodes

C It does not uniquely identify tree structure

D It requires extra memory

Inorder traversal alone cannot reconstruct the tree uniquely because different trees can have the same inorder sequence.

Explain how you would serialize a binary tree using preorder traversal including null markers.

Think about visiting nodes root-left-right and marking empty spots.

Describe the steps to deserialize a serialized string back into the original binary tree.

Use the order of values and null markers to rebuild the tree step by step.

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