Solution Approach

The solution uses Level Order Traversal (BFS) to serialize and deserialize the binary tree. Here's how each part works:

Serialization Process

1. Initialize a queue with the root node and an empty result list
2. Process nodes level by level:
   • Dequeue a node from the front
   • If the node exists, append its value to the result and enqueue both its children (even if they're None)
   • If the node is None, append "#" as a placeholder
3. Join all values with commas to create the final serialized string

q = deque([root]) ans = [] while q:  node = q.popleft()  if node:  ans.append(str(node.val))  q.append(node.left) # Add even if None  q.append(node.right) # Add even if None  else:  ans.append("#") return ",".join(ans)

Deserialization Process

1. Split the string by commas to get an array of values
2. Create the root node from the first value
3. Use a queue to track nodes that need children assigned
4. Process values in pairs (left child, right child):
   • For each parent node in the queue
   • Read the next two values from the array
   • If a value is not "#", create a new node and add it to the queue
   • Assign the nodes as left and right children

vals = data.split(",") root = TreeNode(int(vals[0])) q = deque([root]) i = 1 while q:  node = q.popleft()  # Process left child  if vals[i] != "#":  node.left = TreeNode(int(vals[i]))  q.append(node.left)  i += 1  # Process right child  if vals[i] != "#":  node.right = TreeNode(int(vals[i]))  q.append(node.right)  i += 1

Key Data Structures

• Queue (deque): Used for level-order traversal in both serialization and deserialization
• List: Temporarily stores node values during serialization
• String: Final serialized format with comma-separated values

Why This Works

The algorithm maintains a one-to-one correspondence between the serialization and deserialization process:

• During serialization, we process parent nodes before children and record nulls
• During deserialization, we read values in the same order and reconstruct relationships
• The queue ensures we process nodes level by level in both operations
• The "#" markers preserve the tree structure by indicating where nodes are missing

This approach has O(n) time complexity for both operations (visiting each node once) and O(n) space complexity (for the queue and result string).

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider this binary tree:

 1  / \  2 3  / 4

Serialization Process:

1. Initialize: Queue = [1], Result = []

2. Level 0 (root):

   • Dequeue node 1
   • Add "1" to result → Result = ["1"]
   • Enqueue its children → Queue = [2, 3]

3. Level 1:

   • Dequeue node 2

   • Add "2" to result → Result = ["1", "2"]

   • Enqueue its children → Queue = [3, 4, None]

   • Dequeue node 3

   • Add "3" to result → Result = ["1", "2", "3"]

   • Enqueue its children → Queue = [4, None, None, None]

4. Level 2:

   • Dequeue node 4

   • Add "4" to result → Result = ["1", "2", "3", "4"]

   • Enqueue its children → Queue = [None, None, None, None, None]

   • Dequeue None (from node 2's right)

   • Add "#" to result → Result = ["1", "2", "3", "4", "#"]

   • No children to enqueue → Queue = [None, None, None, None]

5. Continue processing remaining Nones:

   • Result = ["1", "2", "3", "4", "#", "#", "#", "#", "#"]

6. Final string: "1,2,3,4,#,#,#,#,#"

Deserialization Process:

1. Split string: vals = ["1", "2", "3", "4", "#", "#", "#", "#", "#"]

2. Create root: node(1), Queue = [node(1)], index i = 1

3. Process node(1)'s children:

   • vals[1] = "2" → Create node(2) as left child
   • vals[2] = "3" → Create node(3) as right child
   • Queue = [node(2), node(3)], i = 3

4. Process node(2)'s children:

   • vals[3] = "4" → Create node(4) as left child
   • vals[4] = "#" → No right child
   • Queue = [node(3), node(4)], i = 5

5. Process node(3)'s children:

   • vals[5] = "#" → No left child
   • vals[6] = "#" → No right child
   • Queue = [node(4)], i = 7

6. Process node(4)'s children:

   • vals[7] = "#" → No left child
   • vals[8] = "#" → No right child
   • Queue = [], i = 9

7. Result: The original tree is reconstructed:

 1  / \  2 3  / 4

The key insight is that we process nodes in the same order during both serialization and deserialization, maintaining the parent-child relationships through the level-order traversal pattern.

Time and Space Complexity

Time Complexity: O(n)

For the serialize method:

• We traverse each node in the tree exactly once using BFS (level-order traversal)
• For each node, we perform constant time operations: appending to the queue, appending to the result list, and string conversion
• The final join operation takes O(m) where m is the total number of elements in the list, which is at most 2n + 1 (including null nodes)
• Overall: O(n)

For the deserialize method:

• We process each value in the serialized string exactly once
• For each non-null value, we create a node and add it to the queue in constant time
• The split operation takes O(k) where k is the length of the string, which is proportional to n
• Overall: O(n)

Space Complexity: O(n)

For the serialize method:

• The queue stores at most O(n) nodes (in the worst case of a complete binary tree, the last level contains approximately n/2 nodes)
• The result list ans stores all nodes plus null markers, which is O(n)
• The final string representation also takes O(n) space
• Overall: O(n)

For the deserialize method:

• The vals array after splitting stores O(n) elements
• The queue stores at most O(n) nodes
• The reconstructed tree itself contains n nodes, taking O(n) space
• Overall: O(n)

Where n is the number of nodes in the binary tree.

Common Pitfalls

1. Not Handling Empty Tree Edge Case

A frequent mistake is forgetting to handle the case when the input tree is None/empty. This causes errors during serialization or deserialization.

Problem Code:

def serialize(self, root):  queue = deque([root]) # Will add None to queue  result = []  while queue:  node = queue.popleft()  result.append(str(node.val)) # Error: None has no attribute 'val'

Solution: Always check for empty tree at the beginning of both methods:

def serialize(self, root):  if root is None:  return ""  # ... rest of the code  def deserialize(self, data):  if not data:  return None  # ... rest of the code

2. Index Out of Bounds in Deserialization

When deserializing, accessing array indices without proper bounds checking can cause IndexError, especially with trailing null nodes.

Problem Code:

while queue:  node = queue.popleft()  # Might exceed array bounds  if values[index] != "#": # IndexError if index >= len(values)  node.left = TreeNode(int(values[index]))  index += 1

Solution: Add bounds checking or ensure serialization doesn't include unnecessary trailing nulls:

while queue and index < len(values):  node = queue.popleft()  if index < len(values) and values[index] != "#":  node.left = TreeNode(int(values[index]))  index += 1  if index < len(values) and values[index] != "#":  node.right = TreeNode(int(values[index]))  index += 1

3. Forgetting to Add None Children to Queue During Serialization

A critical mistake is only adding non-null children to the queue during serialization, which breaks the position mapping needed for deserialization.

Problem Code:

def serialize(self, root):  while queue:  node = queue.popleft()  if node:  result.append(str(node.val))  if node.left: # Wrong: skips None nodes  queue.append(node.left)  if node.right: # Wrong: skips None nodes  queue.append(node.right)

Solution: Always add both children to the queue, even if they're None:

if node:  result.append(str(node.val))  queue.append(node.left) # Add even if None  queue.append(node.right) # Add even if None else:  result.append("#")

4. Using Wrong Delimiter or Conflicting with Node Values

If node values can contain the delimiter character (e.g., negative numbers with "-" delimiter), the deserialization will fail.

Problem Code:

# If using space as delimiter but values might contain spaces return " ".join(result) # Problematic if values have spaces  # Or not handling negative numbers properly values = data.split("-") # Will split "-5" incorrectly

Solution: Use a delimiter that won't appear in node values (comma is safe for integers):

return ",".join(result) # Safe for integer values # For more complex data, consider using JSON or escape sequences

5. Inefficient String Concatenation

Building the serialized string with repeated concatenation instead of joining a list causes O(n²) time complexity.

Problem Code:

def serialize(self, root):  result = ""  while queue:  node = queue.popleft()  if node:  result += str(node.val) + "," # Inefficient string concatenation  else:  result += "#,"

Solution: Collect values in a list and join once at the end:

result = [] while queue:  # ... append to list  result.append(str(node.val)) return ",".join(result) # Single join operation