428. Serialize and Deserialize N-ary Tree

Reference: LeetCode
Difficulty: Hard

Problem

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize an N-ary tree. An N-ary tree is a rooted tree in which each node has no more than N children. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that an N-ary tree can be serialized to a string and this string can be deserialized to the original tree structure.

For example, you may serialize the following 3-ary tree:

as [1 [3[5 6] 2 4]]. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note:

• N is in the range of [1, 1000].
• Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Example:

You may serialize the following tree:
    1
   / \
  2   3
     / \
    4   5
as "[1,2,3,null,null,4,5]"

Clarification: The above format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Analysis

BFS

Note:

Time: $O(N)$ if using StringBuffer for string concatenation.
Space: $O(N)$ since we need to keep the entire tree.

Serialize:

public String serialize(Node root) {
  if (root == null) {
    return null;
  }
  StringBuffer sb = new StringBuffer();
  serializeHelper(root, sb);
  sb.setLength(sb.length() - 1);
  return sb.toString();
}


private void serializeHelper(Node root, StringBuffer sb) {
  if (root == null) {
    return;
  }
  Queue<Node> queue = new LinkedList<>();
  queue.offer(root);

  while (queue.size() > 0) {
    int childSize = queue.size();
    for (int i = 0; i < childSize; ++i) {
      Node p = queue.poll(); // assume not null
      sb.append(p.val + ",");
      sb.append(p.children.size() + ",");
      for (Node child : p.children) {
        queue.offer(child);
      }
    }
  }
}

Deserialize:

public Node deserialize(String data) {
  if (data == null || data.length() == 0) {
    return null;
  }
  
  String[] dataArray = data.split(",");
  List<String> dataList = new ArrayList<>(Arrays.asList(dataArray));

  int index = 0;
  Queue<Node> queue = new LinkedList<>();
  Queue<Integer> sizeQueue = new LinkedList<>();
  Node root = new Node(Integer.valueOf(dataList.get(index)));
  queue.offer(root);
  index += 1;
  sizeQueue.offer(Integer.valueOf(dataList.get(index)));
  index += 1;

  while (queue.size() > 0) {
    int childSize = queue.size();
    for (int i = 0; i < childSize; ++i) {
      Node curr = queue.poll(); // not null
      int currChildSize = sizeQueue.poll();
      curr.children = new ArrayList<>(currChildSize);
      for (int j = 0; j < currChildSize; ++j) {
        int nodeVal = Integer.valueOf(dataList.get(index));
        index += 1;
        int nodeSize = Integer.valueOf(dataList.get(index));
        index += 1;
        Node child = new Node(nodeVal);
        curr.children.add(child);
        queue.offer(child);
        sizeQueue.offer(nodeSize);
      }
    }
  }

  return root;
  
}

DFS (Solution)

Time: $O(N)$ if using StringBuffer for string concatenation.
Space: $O(N)$ since we need to keep the entire tree.

Serialize:

public String serialize(Node root) {
  if (root == null) {
    return null;
  }
  StringBuffer sb = new StringBuffer();
  rserialize(root, sb);
  
  return sb.toString();
}


private void rserialize(Node root, StringBuffer sb) {
  // No base case is required
  sb.append(root.val + ",");
  sb.append(root.children.size() + ","); // record the size of children
  
  for (Node node : root.children) { // when children.size() == 0, it will return
    rserialize(node, sb);
  }
}

Deserialize:

public Node deserialize(String data) {
  if (data == null || data.length() == 0) {
    return null;
  }
  String[] dataArray = data.split(",");
  List<String> dataList = new LinkedList<>(Arrays.asList(dataArray));
  return rdeserialize(dataList);
}

private Node rdeserialize(List<String> L) {
  if (L == null || L.size() == 0) {
    return null;
  }    
  Node root = new Node(Integer.valueOf(L.get(0)));
  L.remove(0);
  int childSize = Integer.valueOf(L.get(0));
  L.remove(0);
  root.children = new ArrayList<>();
  for (int i = 0; i < childSize; ++i) {
    root.children.add(rdeserialize(L));
  }
  return root;
}