Serialize and Deserialize Binary Tree Interview Question for Google

## Serialize and Deserialize a Binary Tree

### Problem Description

The problem requires designing an algorithm to serialize a binary tree into a string representation and subsequently deserialize the string back into the original binary tree structure. The serialization and deserialization process must preserve the structure and content of the original binary tree.

**Clarification:** The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

**Example 1:**

Input: root = [1,2,3,null,null,4,5] Output: [1,2,3,null,null,4,5]


**Example 2:**

Input: root = [] Output: []


### Naive Approach (Breadth-First Search)

A simple way to serialize and deserialize a binary tree is by performing a breadth-first search (BFS). During serialization, we traverse the tree level by level, adding each node's value to a list. If a node is `null`, we add a special marker (e.g., "null") to the list.  During deserialization, we reconstruct the tree from the list, level by level.

#### Code (Python)

```python
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Codec:

    def serialize(self, root):
        if not root:
            return "[]"

        queue = [root]
        result = []

        while queue:
            node = queue.pop(0)
            if node:
                result.append(str(node.val))
                queue.append(node.left)
                queue.append(node.right)
            else:
                result.append("null")

        # Remove trailing nulls
        while result and result[-1] == "null":
            result.pop()

        return '[' + ','.join(result) + ']'

    def deserialize(self, data):
        if data == "[]":
            return None

        data = data[1:-1].split(',')
        root = TreeNode(int(data[0]))
        queue = [root]
        i = 1

        while queue:
            node = queue.pop(0)

            if i < len(data) and data[i] != "null":
                node.left = TreeNode(int(data[i]))
                queue.append(node.left)
            i += 1

            if i < len(data) and data[i] != "null":
                node.right = TreeNode(int(data[i]))
                queue.append(node.right)
            i += 1

        return root

Optimal Approach (Preorder Traversal)

Another approach is to use preorder traversal. We can serialize the tree by visiting the root first, then the left subtree, and then the right subtree. We use a special marker for null nodes. When deserializing, we reconstruct the tree using the preorder traversal sequence.

Code (Python)

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Codec:

    def serialize(self, root):
        def preorder(node):
            if not node:
                result.append('null')
                return
            result.append(str(node.val))
            preorder(node.left)
            preorder(node.right)

        result = []
        preorder(root)
        return ','.join(result)

    def deserialize(self, data):
        def buildTree():
            val = next(data_iter)
            if val == 'null':
                return None
            node = TreeNode(int(val))
            node.left = buildTree()
            node.right = buildTree()
            return node

        data_iter = iter(data.split(','))
        return buildTree()

Big O Analysis

BFS Approach:

Time Complexity:
- Serialization: O(N), where N is the number of nodes in the tree because we visit each node once.
- Deserialization: O(N), where N is the number of nodes in the tree because we reconstruct each node once.
Space Complexity:
- Serialization: O(N), where N is the number of nodes in the tree due to the queue used for BFS and the list to store the serialized data.
- Deserialization: O(N), where N is the number of nodes in the tree due to the queue used for reconstruction and the list to store the deserialized data.

Preorder Traversal Approach:

Time Complexity:
- Serialization: O(N), where N is the number of nodes in the tree because we visit each node once.
- Deserialization: O(N), where N is the number of nodes in the tree because we reconstruct each node once.
Space Complexity:
- Serialization: O(N), where N is the number of nodes in the tree in the worst case (skewed tree) due to the recursion stack.
- Deserialization: O(N), where N is the number of nodes in the tree in the worst case (skewed tree) due to the recursion stack.

Edge Cases

Empty Tree: Both algorithms correctly handle an empty tree. The serialization of an empty tree is "[]" for the BFS approach and "null" for the Preorder approach. The deserialization correctly returns None.
Tree with Only Root: Both algorithms handle a tree with only a root node correctly. The serialization will be "[root.val]" for BFS and "root.val" for Preorder.
Skewed Tree (Left or Right): Both algorithms handle skewed trees correctly, though the space complexity of the Preorder approach might be significant for very skewed trees.
Large Trees: The main limitation for extremely large trees will be memory. For BFS, the queue and list used to store the tree can become very large. For Preorder, the recursion stack can become very deep.

Key Differences and Considerations

BFS vs. Preorder: BFS is generally easier to understand and implement, but it may use slightly more memory because it stores each level in the queue. Preorder is more memory-efficient in balanced trees, but its space complexity becomes O(N) in skewed trees due to the recursion depth.
Error Handling: The provided code does not explicitly handle malformed input during deserialization (e.g., non-integer values where integers are expected). Additional error handling could be added to make the solution more robust.
Choice of Marker: The choice of the marker for null nodes (e.g., "null") is arbitrary. It could be any value that is not a valid node value.