Serialize and Deserialize Binary Tree

Problem Statement

Serialization involves converting a data structure or object into a sequence of bits, enabling it to be stored efficiently in a file or memory buffer or transmitted seamlessly over a network to be accurately reconstructed on the same or a different computer environment. This task asks for designing algorithms for both serialization and deserialization of a binary tree. Serialization would transform the binary tree into a string while deserialization would do the reverse, converting the string back to the original binary tree structure. While there's freedom in choosing the method of serialization and deserialization, the crux lies in ensuring that the final deserialized tree mirrors the original structure entirely.

Examples

Example 1

Input:

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

Output:

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

Example 2

Input:

root = []

Output:

[]

Constraints

• The number of nodes in the tree is in the range [0, 104].
• -1000 <= Node.val <= 1000

Approach and Intuition

Given the nature of binary trees, and the challenge to serialize and deserialize them, one could use several approaches. Below are instinctive reasoning and methodical steps for potential approaches, focusing on the examples provided:

1. Preorder Traversal (Recursive Approach):

   • Serialize the binary tree using preorder traversal (visit the root, then left, and right child). Represent null pointers with a specific character or symbol, say 'X', to denote the absence of a node.
   • To deserialize, read the string and reconstruct the tree by creating nodes for non-null symbols and skipping creation for 'X' or null equivalent, recursively setting left and right children.

2. Level Order Traversal (BFS-Approach):

   • Use a queue to perform a breadth-first search. Serialize nodes level by level. For each node, if it's not null, record its value; if null, record a special symbol.
   • For deserialization, use a queue to maintain the order of node creation. For each non-null value, create a node and attach left and right children accordingly, maintaining their order in the queue.

3. Parenthetical Representation (Advanced Approach):

   • Utilize a method reminiscent of Lisp syntax: serialize each node with its left and right children enclosed in parentheses. For example, a node with left and right children might be represented as ( value ( left_subtree ) ( right_subtree ) ), using empty parentheses for null children.
   • Deserialize recursively by parsing the string from left to right, constructing each node and its respective subtrees from each parenthetical group.

Examples Illustration:

• Example 1:

  • For input root = [1,2,3,null,null,4,5], the serialization might look like 1,2,X,X,3,4,X,X,5,X,X using the Preorder approach with 'X' as null.
  • When deserializing 1,2,X,X,3,4,X,X,5,X,X, reconstruct the original tree by creating a root node at '1', then a left child '2' with no further descendants, then a right child '3' with children '4' and '5'.

• Example 2:

  • An empty tree root = [] when serialized would simply be X or an equivalent empty representation.
  • Deserializing this would confirm returning an empty or null root, corroborating the consistency of the approach.

These approaches, adaptable based on specific requirements or constraints, ensure that the binary tree's structural integrity is maintained post deserialization, regardless of the serialization method chosen.

Solutions

public class Codec {
  public TreeNode buildTree(List<String> dataArray) {
    if (dataArray.get(0).equals("null")) {
      dataArray.remove(0);
      return null;
    }

    TreeNode node = new TreeNode(Integer.parseInt(dataArray.get(0)));
    dataArray.remove(0);
    node.left = buildTree(dataArray);
    node.right = buildTree(dataArray);

    return node;
  }

  public TreeNode deserialize(String encodedData) {
    String[] elements = encodedData.split(",");
    List<String> elementList = new LinkedList<>(Arrays.asList(elements));
    return buildTree(elementList);
  }
};