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:

1 2 3 4 5 6 7

You may serialize the following tree: 1 / \ 23 / \ 45 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.

public Node deserialize(String data){ if (data == null || data.length() == 0) { returnnull; } 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.

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

privatevoidrserialize(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); } }