hard Serialize and Deserialize N-ary Tree

Problem Statement

Serialization is the process of converting a data structure or an object into a sequence of bits. This conversion allows the data to be stored in a file, transferred over a network, or saved in memory, with the ability to reconstruct the data later.

Design an algorithm to serialize and deserialize an N-ary tree.

An N-ary tree is a rooted tree where each node can have up to N children. The serialize() method should transform an N-ary tree into a string format and then the deserialize() method should be able to convert this string back to the original tree structure.

There is no restriction on how to perform serialization or deserialization, as long as the process accurately preserves the tree structure.

Examples

Example 1:

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

• Expected Output: [1,null,2,3,4,null,5,6]
• Explanation: We serialize the deserialize the given N-ary tree.

Example 2:

• Input: root = [10,null,20,30,null,40,50,60]

• Expected Output: [10,null,20,30,null,40,50,60]
• Explanation: We serialize the deserialize the given N-ary tree.

Example 3:

• Input: root = [100,null,200,300,400,500,null,600,700,null,800]

• Expected Output: [100,null,200,300,400,500,null,600,700,null,800]

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.

Solution

To solve this problem, we need a method to convert an N-ary tree into a string that reflects its structure and a reverse method to rebuild the original tree from this string. For serialization, a level order traversal method can be effective. We traverse the tree level by level, appending each node's value to a result string, and use a separator like null to indicate the end of each level. This approach ensures that each node's children are placed correctly in the resulting string.

For deserialization, the reverse process is applied. We use a queue to keep track of the nodes and build the tree level by level based on the values read from the string. This approach maintains the original structure and can handle trees with different branching factors. It is effective because it allows us to process each node once, ensuring optimal performance.

Step-by-Step Algorithm

1. Initialize the Serialization Process:

   • Check if the root is null.
     • If root is null, return an empty string "" because there is nothing to serialize.
   • If root is not null, proceed to the next steps.
   • Create an empty StringBuilder object sb to construct the serialized string efficiently.
   • Create a queue (Queue<Node> queue) to perform a level-order traversal starting with the root node.
     • Add the root node to the queue using queue.offer(root).

2. Level Order Traversal for Serialization:

   • Use a while loop to traverse through the nodes as long as the queue is not empty (while (!queue.isEmpty())):
     • Dequeue a node from the front of the queue using Node node = queue.poll().
     • Check if the dequeued node is null:
       • If the node is null, it indicates the end of a level in the tree.
         • Append "null," to sb to mark the end of the level in the serialized string.
         • Use continue to skip the rest of the loop and start the next iteration.
     • If the dequeued node is not null:
       • Append the node's value (node.val) to sb followed by a comma "," to separate values.
       • Enqueue all children of the current node:
         • Iterate over the node's children (for (Node child : node.children)):
           • Use queue.offer(child) to add each child node to the queue for further processing.
       • Mark the end of the current level:
         • Enqueue a null value (queue.offer(null)) to indicate the end of the level in the tree.

3. Finalize the Serialization:

   • Once the queue is empty, all nodes have been processed, and the entire tree is serialized.
   • Convert the StringBuilder object sb to a string and return it.

4. Initialize the Deserialization Process:

   • Check if the input data is empty:
     • If data is empty, return null since there is no tree to deserialize.
   • Split the input string data into an array of strings (String[] nodes) using a comma "," as the delimiter. This gives a list of node values in the order they were serialized.
   • Create a new Node object root with the value parsed from nodes[0] (the first element in the list) and an empty list of children (new ArrayList<>()).
   • Create a queue (Queue<Node> queue) to help in reconstructing the tree level-by-level:
     • Add the root node to the queue using queue.offer(root).

5. Reconstruct the Tree Level-by-Level:

   • Use a for loop to iterate over the nodes array starting from index 1:
     • Get the Next Parent Node:
       • Dequeue a parent node from the queue using Node parent = queue.poll().
     • Process All Children for the Current Parent Node:
       • Use a while loop to add all children of the parent node until a "null" value is encountered:
         • Check if nodes[i] is "null":
           • If not, create a new Node object child with the value parsed from nodes[i] and an empty list of children.
           • Add the child node to the parent node's list of children (parent.children.add(child)).
           • Enqueue the child node to the queue for processing its children in the future (queue.offer(child)).
           • Increment i to move to the next node value.
       • When "null" is found:
         • End processing for the current parent node and move to the next node in the queue.

6. Return the Reconstructed Tree:

   • Once the entire nodes array has been processed, the original tree structure is reconstructed.
   • Return the root node, which now represents the deserialized N-ary tree.

Algorithm Walkthrough

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

• Step 1: Initialize the process.

  • Check if the root is null:
    • If root is null, return an empty string "" because there is nothing to serialize.
    • In our case, root is not null, so proceed.
  • Initialize tools for serialization:
    • Create a StringBuilder sb to build the output string efficiently.
    • Create a queue (queue) to traverse the tree level by level and add the root node to it:
      queue = [1].

• Step 2: Start level-order traversal using the queue.

  • Iteration 1:

    • Dequeue node 1:
      queue = [].
      Append the value 1 to sb:
      sb = "1,".
    • Enqueue children of node 1 (2, 3, 4):
      queue = [2, 3, 4].
    • Mark the end of the level with null:
      queue = [2, 3, 4, null].

  • Iteration 2:

    • Dequeue node 2:
      queue = [3, 4, null].
      Append the value 2 to sb:
      sb = "1,2,".
    • Enqueue children of node 2 (5, 6):
      queue = [3, 4, null, 5, 6].

  • Mark the end of the level with null:
    queue = [3, 4, null, 5, 6, null].

  • Iteration 3:

    • Dequeue node 3:
      queue = [4, null, 5, 6, null].
      Append the value 3 to sb:
      sb = "1,2,3,".
    • Node 3 has no children:
    • Mark the end of the level with null:
      queue = [4, null, 5, 6, null, null].

  • Iteration 4:

    • Dequeue node 4:
      queue = [null, 5, 6, null, null].
      Append the value 4 to sb:
      sb = "1,2,3,4,".
    • Node 4 has no children:
    • Mark the end of the level with null:
      queue = [null, 5, 6, null, null, null].

  • Iteration 5:

    • Dequeue null:
      queue = [5, 6, null, null, null].
      Append "null," to sb to indicate the end of the level:
      sb = "1,2,3,4,null,".

  • Iteration 6:

    • Dequeue node 5:
      queue = [6, null, null, null].
      Append the value 5 to sb:
      sb = "1,2,3,4,null,5,".
    • Node 5 has no children:
    • Mark the end of the level with null:
      queue = [6, null, null, null, null].

  • Iteration 7:

    • Dequeue node 6:
      queue = [null, null, null, null].
      Append the value 6 to sb:
      sb = "1,2,3,4,null,5,6,".
    • Node 6 has no children:
    • Mark the end of the level with null:
      queue = [null, null, null, null, null].

  • Iteration 8, 9, 10, 11, 12:

    • Dequeue all null:
      queue = [].
      Append total 5 "null," to sb to indicate the end of the level:
      sb = "1,2,3,4,null,5,6,null,null,null,null,null,"`.

2. Deserialization Process

We will rebuild the tree from the serialized string "1,2,3,4,null,5,6,null,null,null,null,null,".

• Step 1: Initialize deserialization.

  • Check if the input data is empty:
    • If data is empty, return null since there is no tree to deserialize.
    • In our case, data is not empty, so proceed.
  • Split the input string data into nodes:
    nodes = ["1", "2", "3", "4", "null", "5", "6", "null", "null", "null", "null", "null"].
  • Create the root node from nodes[0]:
    Node root = new Node(1, new ArrayList<>()).
  • Initialize a queue to help build the tree level by level:
    queue = [1].

• Step 2: Reconstruct the tree using the queue.

  • Iteration 1:

    • Dequeue parent node 1:
      queue = [].
    • Process children of node 1:
      • Check nodes[1] (2):
        • It is not "null", so create Node(2) and add to 1's children:
          root.children = [2].
        • Enqueue Node(2) to the queue:
          queue = [2].
      • Check nodes[2] (3):
        • It is not "null", so create Node(3) and add to 1's children:
          root.children = [2, 3].
        • Enqueue Node(3) to the queue:
          queue = [2, 3].
      • Check nodes[3] (4):
        • It is not "null", so create Node(4) and add to 1's children:
          root.children = [2, 3, 4].
        • Enqueue Node(4) to the queue:
          queue = [2, 3, 4].
      • Check nodes[4] (null):
        • It is "null", marking the end of children for node 1.

  • Iteration 2:

    • Dequeue parent node 2:
      queue = [3, 4].
    • Process children of node 2:
      • Check nodes[5] (5):
        • It is not "null", so create Node(5) and add to 2's children:
          node 2.children = [5].
        • Enqueue Node(5) to the queue:
          queue = [3, 4, 5].
      • Check nodes[6] (6):
        • It is not "null", so create Node(6) and add to 2's children:
          node 2.children = [5, 6].
        • Enqueue Node(6) to the queue:
          queue = [3, 4, 5, 6].
      • Check nodes[7] (null):
        • It is "null", marking the end of children for node 2.

  • Iteration 3:

    • Dequeue parent node 3:
      queue = [4, 5, 6].
    • Process children of node 3:
      • Check nodes[8] (null):
        • It is "null", so 3 has no children.

  • Iteration 4:

    • Dequeue parent node 4:
      queue = [5, 6].
    • Process children of node 4:
      • Check nodes[9] (null):
        • It is "null", so 4 has no children.

  • Iteration 5:

    • Dequeue nodes 5 and 6:
      queue = [].
    • Process children of nodes 5 and 6:
      • Both nodes have no children, so continue.

• Result:
  The original tree structure is reconstructed correctly.

Code

import java.util.*;

// class NAryNode {
//     public int val;
//     public List<NAryNode> children;

//     public NAryNode() {}

//     public NAryNode(int _val) {
//         val = _val;
//     }

//     public NAryNode(int _val, List<NAryNode> _children) {
//         val = _val;
//         children = _children;
//     }
// }

public class Solution {

  public String serialize(NAryNode root) {
    if (root == null) return ""; // If the tree is empty, return an empty string
    StringBuilder sb = new StringBuilder();
    Queue<NAryNode> queue = new LinkedList<>();
    queue.offer(root); // Add the root node to the queue

    while (!queue.isEmpty()) {
      NAryNode node = queue.poll(); // Remove the front node from the queue
      if (node == null) {
        sb.append("null,"); // Add "null" to the string to indicate the end of a level
        continue;
      }
      sb.append(node.val).append(","); // Append the value of the current node to the string
      for (NAryNode child : node.children) {
        queue.offer(child); // Add all children of the current node to the queue
      }
      queue.offer(null);
    }
    return sb.toString();
  }

  public NAryNode deserialize(String data) {
    if (data.isEmpty()) return null; // If the input string is empty, return null
    String[] nodes = data.split(","); // Split the string by commas to get node values
    NAryNode root = new NAryNode(Integer.parseInt(nodes[0]), new ArrayList<>());
    Queue<NAryNode> queue = new LinkedList<>();
    queue.offer(root); // Add the root node to the queue

    for (int i = 1; i < nodes.length; i++) {
      NAryNode parent = queue.poll(); // Get the next parent node from the queue
      while (!nodes[i].equals("null")) {
        NAryNode child = new NAryNode(
          Integer.parseInt(nodes[i]),
          new ArrayList<>()
        );

        parent.children.add(child); // Add the child node to the parent
        queue.offer(child); // Add the child to the queue to process its children
        i++;
      }
    }
    return root;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    // Create nodes for the tree
    NAryNode root = new NAryNode(1); // Root node with value 1
    NAryNode node2 = new NAryNode(2); // Node with value 2
    NAryNode node3 = new NAryNode(3); // Node with value 3
    NAryNode node4 = new NAryNode(4); // Node with value 4
    NAryNode node5 = new NAryNode(5); // Node with value 5
    NAryNode node6 = new NAryNode(6); // Node with value 6

    // Build the tree structure
    root.children.add(node2); // Adding node 2 as a child of root
    root.children.add(node3); // Adding node 3 as a child of root
    root.children.add(node4); // Adding node 4 as a child of root

    node2.children.add(node5); // Adding node 5 as a child of node 2
    node2.children.add(node6); // Adding node 6 as a child of node 2

    // Serialize the tree
    String serializedData = solution.serialize(root);
    System.out.println("Serialized Output: " + serializedData); // Output should be: [1,null,2,3,4,null,5,6]

    // Deserialize the tree back from the serialized data
    NAryNode deserializedRoot = solution.deserialize(serializedData);
    System.out.println(
      "Deserialized Output: " + solution.serialize(deserializedRoot)
    ); // Should match the serialized output
  }
}

Complexity Analysis

Time Complexity

The time complexity for both serialization and deserialization is , where N is the number of nodes in the tree. This is because each node is visited once during serialization and deserialization.

Space Complexity

The space complexity is as well, mainly due to the additional storage needed for the queue during the traversal.