Serialize and Deserialize Binary Tree in DSA Typescript

Mental Model

We turn a tree into a string to save or send it, then rebuild the exact same tree from that string.

Analogy: Like writing a family tree on paper and later using that paper to redraw the same family tree exactly as it was.

    1
   / \
  2   3
     / \
    4   5

Serialize as: "1,2,null,null,3,4,null,null,5,null,null,"

Dry Run Walkthrough

Input: Binary tree: 1 with left child 2 and right child 3; 3 has children 4 and 5

Goal: Convert the tree into a string and then rebuild the same tree from that string

Step 1: Start serialization at root node 1

Serialize: "1,"

Why: We record the root value first to start the string

Step 2: Serialize left subtree of 1 (node 2)

Serialize: "1,2,"

Why: We record left child value next

Step 3: Serialize left child of 2 which is null

Serialize: "1,2,null,"

Why: We mark null to remember no left child

Step 4: Serialize right child of 2 which is null

Serialize: "1,2,null,null,"

Why: We mark null to remember no right child

Step 5: Serialize right subtree of 1 (node 3)

Serialize: "1,2,null,null,3,"

Why: We record right child value after left subtree

Step 6: Serialize left child of 3 (node 4)

Serialize: "1,2,null,null,3,4,"

Why: We record left child of 3

Step 7: Serialize left and right children of 4 which are null

Serialize: "1,2,null,null,3,4,null,null,"

Why: Mark nulls for 4's children

Step 8: Serialize right child of 3 (node 5) and its null children

Serialize: "1,2,null,null,3,4,null,null,5,null,null,"

Why: Complete serialization with right subtree

Step 9: Start deserialization from string

Deserialize: reading "1" creates root node 1

Why: We rebuild root first

Step 10: Deserialize left subtree from next tokens "2,null,null"

Deserialize: node 2 with null children attached to root

Why: Rebuild left subtree exactly

Step 11: Deserialize right subtree from tokens "3,4,null,null,5,null,null"

Deserialize: node 3 with children 4 and 5 rebuilt

Why: Rebuild right subtree exactly

Result:

Final tree rebuilt exactly as original:
    1
   / \
  2   3
     / \
    4   5

Annotated Code

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

class Codec {
  // Serialize tree to string
  serialize(root: TreeNode | null): string {
    if (root === null) return "null,"; // mark null nodes
    return root.val + "," + this.serialize(root.left) + this.serialize(root.right);
  }

  // Deserialize string to tree
  deserialize(data: string): TreeNode | null {
    const nodes = data.split(",");
    function helper(): TreeNode | null {
      const val = nodes.shift();
      if (val === "null" || val === undefined || val === "") return null; // null node found or empty string
      const node = new TreeNode(parseInt(val));
      node.left = helper(); // build left subtree
      node.right = helper(); // build right subtree
      return node;
    }
    return helper();
  }
}

// Driver code
const root = new TreeNode(1, new TreeNode(2), new TreeNode(3, new TreeNode(4), new TreeNode(5)));
const codec = new Codec();
const serialized = codec.serialize(root);
console.log("Serialized:", serialized);
const deserializedRoot = codec.deserialize(serialized);

// Helper to print tree inorder for verification
function inorderPrint(node: TreeNode | null): string {
  if (!node) return "null";
  return node.val + "(" + inorderPrint(node.left) + "," + inorderPrint(node.right) + ")";
}
console.log("Deserialized inorder:", inorderPrint(deserializedRoot));

if (root === null) return "null,";

mark null nodes explicitly to preserve tree shape

return root.val + "," + this.serialize(root.left) + this.serialize(root.right);

serialize current node then left and right subtrees recursively

const val = nodes.shift();

read next value from serialized data to rebuild node

if (val === "null" || val === undefined) return null;

detect null marker to stop recursion on this branch

node.left = helper();

rebuild left subtree recursively

node.right = helper();

rebuild right subtree recursively

Output

Serialized: 1,2,null,null,3,4,null,null,5,null,null,

Deserialized inorder: 1(2(null,null),3(4(null,null),5(null,null)))

Complexity Analysis

Time: O(n) because we visit each node once during serialization and once during deserialization

Space: O(n) for the recursion stack and the serialized string storing all nodes

vs Alternative: Compared to level-order serialization, preorder with null markers is simpler and uses recursion, but both have similar O(n) cost

Edge Cases

Empty tree (root is null)

Serialization returns "null," and deserialization returns null tree

DSA Typescript

if (root === null) return "null,";

Tree with only one node

Serialization returns "val,null,null," and deserialization rebuilds single node

DSA Typescript

if (val === "null" || val === undefined) return null;

Tree with only left or only right children

Null markers preserve missing children so tree shape is exact

DSA Typescript

if (val === "null" || val === undefined) return null;

When to Use This Pattern

When you need to save or send a tree structure and later rebuild it exactly, use serialize/deserialize with preorder traversal and null markers to preserve shape.

Common Mistakes

Mistake: Not marking null children during serialization, causing incorrect tree shape on deserialization
Fix: Always serialize null children explicitly as "null," to keep tree structure

Mistake: Using split without handling trailing commas causing empty strings in array
Fix: Ensure to handle or ignore empty strings after splitting serialized data

Summary

It converts a binary tree to a string and back to the same tree.

Use it when you want to save or transfer a tree and restore it later exactly.

The key is to record null children explicitly to keep the tree shape intact.