Serialize and Deserialize N-ary Tree

Problem Statement

Serialization in computing involves converting a data structure or object state into a format that can be stored or transmitted and later reconstructed. This challenge requires designing an algorithm capable of both serializing and deserializing an N-ary tree, where the tree consists of nodes that each have up to N children. The pivotal requirement is that the serialized N-ary tree, represented as a string, can be transformed back into the original tree structure with appropriate consistency. The solution should adhere strictly to being stateless, not relying on any external or static storage to maintain the state during encoding or decoding.

Examples

Example 1

Input:

root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Output:

[1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Example 2

Input:

root = [1,null,3,2,4,null,5,6]

Output:

[1,null,3,2,4,null,5,6]

Example 3

Input:

root = []

Output:

[]

Constraints

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000
• Do not use class member/global/static variables to store states. Your encode and decode algorithms should be stateless.

Approach and Intuition

When tackling the serialization and deserialization of an N-ary tree, the main consideration is how to record the relationship between nodes and their children efficiently and clearly. Let's break down the approach:

1. Serialization Approach:

   • For each node, record the node's value followed by a special marker (e.g., null) that indicates the end of children for that node.
   • Proceed recursively or iteratively with each child of the current node, appending their values to the serialization string.
   • Ensure each node's children are denoted clearly such that when deserializing, the exact tree structure can be reconstructed.

2. Deserialization Approach:

   • Parse the serialized string from left to right.
   • Use a stack or queue to help rebuild the tree structure. Each time a node value is read, create a new tree node.
   • Upon encountering the special marker, determine the closure of a node's children list and attach the popped nodes to their parent from the last node in the stack.

The examples provided utilize a specific serialized pattern:

• Example 1:

  • Series of node values interspersed with null to denote the transition to a new set of children or moving up the tree hierarchy.
  • From the given serialization pattern, reconstruct the tree by assigning children appropriately based on the null markers.

• Example 2 and Example 3:

  • Similar handling where null separates nodes or indicates the absence of further nodes.

The constraints of the problem ensure that the solution needs to be efficient in both space and execution time, where maximum node and depth levels are stipulated. This fosters an approach that must be both lightweight concerning memory usage and optimized for speed to handle potentially large trees effectively.

Solutions

class TreeSerializer {

    // Transforms a tree to a unified string.
    public String encode(Node root) {
        if (root == null) {
            return "";
        }

        StringBuilder result = new StringBuilder();
        constructSerialization(root, result);
        return result.toString();
    }

    private void constructSerialization(Node root, StringBuilder result) {

        Queue<Node> queue = new LinkedList<>();
        Node levelTerminator = new Node();
        Node childrenSeparator = new Node();
        queue.add(root);
        queue.add(levelTerminator);

        while (!queue.isEmpty()) {

            Node currentNode = queue.poll();

            if (currentNode == levelTerminator) {
                result.append('#');
                if (!queue.isEmpty()) {
                    queue.add(levelTerminator);
                }
            } else if (currentNode == childrenSeparator) {
                result.append('`');
            } else {
                result.append((char) (currentNode.val + '0'));
                for (Node child : currentNode.children) {
                    queue.add(child);
                }
                if (queue.peek() != levelTerminator) {
                    queue.add(childrenSeparator);
                }
            }
        }
    }

    // Rebuilds the tree from the encoded data.
    public Node decode(String data) {
        if (data.isEmpty()) {
            return null;
        }

        Node rootNode = new Node(data.charAt(0) - '0', new ArrayList<>());
        rebuildTree(data, rootNode);
        return rootNode;
    }

    private void rebuildTree(String data, Node root) {

        LinkedList<Node> thisLevel = new LinkedList<>();
        LinkedList<Node> previousLevel = new LinkedList<>();
        thisLevel.add(root);
        Node currentParent = root;

        for (int i = 1; i < data.length(); i++) {
            char charData = data.charAt(i);
            if (charData == '#') {
                previousLevel = thisLevel;
                thisLevel = new LinkedList<>();
                currentParent = previousLevel.poll();
            } else {
                if (charData == '`') {
                    currentParent = previousLevel.poll();
                } else {
                    Node childNode = new Node(charData - '0', new ArrayList<>());
                    thisLevel.add(childNode);
                    currentParent.children.add(childNode);
                }
            }
        }
    }
}