Problem 3

On to the next!

❓ PROMPT

Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree.

For example, given the following Node class

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

The following test should pass:

node = Node('root', Node('left', Node('left.left')), Node('right'))
assert deserialize(serialize(node)).left.left.val == 'left.left'

🔍 EXPLORE

Alright, let's break this down

Serialization is the process of converting an object or data structure into a format that can be easily stored or transmitted, and later reconstructed. In this case, we're turning a binary tree into a string.

Deserialization, on the other hand, is the reverse process. We're taking that string and turning it back into the original binary tree.

For our challenge, we have a binary tree with nodes that can have a value, a left child, and a right child. The goal is to create two functions:

1. serialize(root): Converts the tree into a string.
2. deserialize(s): Converts the string back into the tree.

The provided test case gives us a tree with a root node, a left child, a right child, and a left-left grandchild. After serializing and then deserializing the tree, the value of the left-left grandchild should still be 'left.left'.

Assumptions and Discoveries:

1. The tree nodes contain string values.
2. The tree can be of any depth and can be unbalanced.
3. We need a way to represent null or missing children in the serialized format.

Edge/Corner Cases:

1. What if the tree is empty (i.e., the root is None)?
2. What if the tree has only one node?
3. What if the tree is heavily unbalanced, e.g., all nodes are only on the left or only on the right?

🗣️ CLARIFYING QUESTIONS:

The prompt seems clear, but just to be sure: Is there a preferred format for the serialized string? And are there any constraints on the size or depth of the tree?

💡 BRAINSTORM:

There are several ways to serialize a binary tree:

Remember, the key to successful serialization and deserialization is not just the order in which nodes are visited, but also the ability to accurately represent the tree's structure in a linear format. This often requires including placeholders or markers for null values or other structural elements. For our purposes, Pre-order Traversal seems the most straightforward. We can represent null children with a special character, say #, and separate node values with a delimiter, like ,.

Time and Space Complexity:

• serialize: O(n) time, where n is the number of nodes. O(n) space for the resulting string.
• deserialize: O(n) time to reconstruct the tree and O(n) space for the tree itself.

📝 PLAN:

1. For serialize:

• Start at the root.
• If the node is None, append # to the result string.
• Otherwise, append the node value, followed by a delimiter.
• Recursively serialize the left and right children.

2. For deserialize:

• Split the string by the delimiter to get a list of values.
• Use a list iterator or index to keep track of the current value.
• If the current value is #, return None.
• Otherwise, create a new node with the current value.
• Recursively deserialize the left and right children.
• Return the constructed node.

🛠️ IMPLEMENT

They gave us the node class in Python so let's make use of it

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

def serialize(root):
    if not root:
        return '#'
    return root.val + ',' + serialize(root.left) + ',' + serialize(root.right)

def deserialize(data):
    def helper(values):
        val = next(values)
        if val == '#':
            return None
        node = Node(val)
        node.left = helper(values)
        node.right = helper(values)
        return node

    values = iter(data.split(','))
    return helper(values)

🔄 REVIEW AND REFACTOR:

The code seems clean and adheres to the plan. The serialization uses pre-order traversal, and the deserialization reconstructs the tree in the same order. The use of an iterator (values) in the deserialize function ensures we're processing each value from the serialized string only once.

📊 ANALYZE:

The final solution has a time and space complexity of O(n) for both serialization and deserialization, where n is the number of nodes in the tree. This is because we're visiting each node exactly once.

🎁 BONUS:

What if instead of a binary tree we're given a linked list that can potentially contain cycles. Apple asked me this question once and I bombed it.

Assumptions and Discoveries:

• The linked list nodes contain string values.
• The linked list can have cycles.
• We'll use UUIDs to uniquely identify nodes and detect cycles.

Edge/Corner Cases:

• What if the linked list is empty?
• What if the linked list has only one node?
• What if the linked list is a complete cycle?

💡 BRAINSTORM:

For serialization:

1. Traverse the linked list.
2. For each node, store its UUID, value, and the UUID of the next node.
3. If we encounter a UUID that we've seen before, we've detected a cycle and can stop the serialization.

For deserialization:

1. Read the serialized data and reconstruct the linked list.
2. Use a dictionary to map UUIDs to nodes so we can easily connect nodes.

Time and Space Complexity:

• serialize: O(n) time, where n is the number of nodes. O(n) space for the resulting string.
• deserialize: O(n) time to reconstruct the linked list and O(n) space for the linked list itself.

📝 PLAN:

For serialize:

1. Traverse the linked list.
2. For each node, append its UUID, value, and the UUID of the next node to the result string.
3. If we encounter a UUID we've seen before, stop the traversal.

For deserialize:

1. Split the string to get the list of UUIDs and values.
2. Use a dictionary to map UUIDs to nodes.
3. Reconstruct the linked list using the dictionary.

class Node:
    def __init__(self, val, next=None):
        self.uuid = str(uuid.uuid4())
        self.val = val
        self.next = next

def serialize(head):
    visited = set()
    result = []
    current = head
    while current:
        if current.uuid in visited:
            result.append(current.uuid)
            break
        visited.add(current.uuid)
        result.append(current.uuid + '|' + current.val)
        current = current.next
    return ','.join(result)

def deserialize(data):
    if not data:
        return None
    values = data.split(',')
    uuid_to_node = {}
    prev_node = None
    for val in values:
        if '|' in val:
            node_uuid, node_val = val.split('|')
            node = Node(node_val)
            node.uuid = node_uuid
            uuid_to_node[node_uuid] = node
            if prev_node:
                prev_node.next = node
            else:
                head = node
            prev_node = node
        else:
            # This is a cycle
            prev_node.next = uuid_to_node[val]
            break
    return head

🧪 VERIFY:

node3 = Node('three')
node2 = Node('two', node3)
node1 = Node('one', node2)
node3.next = node1  # Create a cycle
serialized_data = serialize(node1)
deserialized_list = deserialize(serialized_data)
assert deserialized_list.val == 'one'
assert deserialized_list.next.next.next.val == 'one'

🔄 REVIEW AND REFACTOR:

Ah, the joys of revisiting code. It's like looking at your past self and wondering, "What was I thinking?" But, surprisingly, this one doesn't look half bad. The serialization and deserialization functions are straightforward, and the use of UUIDs to detect cycles is pretty clever if I do say so myself. Maybe I've learned a thing or two since that Apple interview debacle.

However, a potential improvement could be optimizing the space used by the serialized string. Right now, we're storing UUIDs for every node, which can get lengthy. But, considering the need to handle cycles, it's a trade-off I'm willing to make.

📊 ANALYZE:

The solution's time complexity is O(n) for both serialization and deserialization, where n is the number of nodes in the linked list. This is because we're visiting each node exactly once. The space complexity is also O(n) due to the storage of UUIDs and values in the serialized string.

But let's be real. In a world where we have terabytes of storage in our pockets, are we really going to quibble over a few extra bytes for UUIDs?

📚 FINAL THOUGHTS:

Ah, the infamous "lessons learned" section. Well, here's a lesson for you: Always be prepared for the unexpected in coding interviews. Just when you think you've nailed it, they'll throw a curveball like handling cycles in a linked list. Trust me, I've been there.

But on the bright side, every bombed interview is an opportunity to learn and grow. And maybe, just maybe, write a post to vent a little. So, here's to the never-ending journey of learning, improving, and occasionally facepalming at our past mistakes. Cheers! 🍻