297. Serialize and Deserialize Binary Tree (Hard)

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 a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary 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 tree

    1
   / \
  2   3
     / \
    4   5

as"[1,2,3,null,null,4,5]", just the same ashow LeetCode OJ 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.

Solution 1: BFS O(n);O(w) O(n);O(n)

Use BFS to traverse the tree in level order and use string "#"(pound) to represent null nodes.
When deserialize the string, assign left and right child for each non-null node, and add the non-null children to the queue, which will be handled later.

    public String serialize(TreeNode root) { //O(n);O(w)
        if (root == null) {
            return "";
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        StringBuilder res = new StringBuilder();
        while (!queue.isEmpty()) {
            TreeNode head = queue.poll();
            if (head == null) {
                res.append("#,");
                continue;
            }
            res.append(head.val + ",");
            queue.offer(head.left);
            queue.offer(head.right);
        }
        return res.toString();
    }

    public TreeNode deserialize(String data) { //O(n);O(n)
        if (data == "") {
            return null;
        }

        String[] values = data.split(",");
        TreeNode root = new TreeNode(Integer.parseInt(values[0]));
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        for (int i = 1; i < values.length; i++) {
            TreeNode parent = queue.poll();
            if (!values[i].equals("#")) {
                TreeNode left = new TreeNode(Integer.parseInt(values[i]));
                parent.left = left;
                queue.offer(left);
            }
            if (!values[++i].equals("#")) {
                TreeNode right = new TreeNode(Integer.parseInt(values[i]));
                parent.right = right;
                queue.offer(right);
            }
        }
        return root;
    }

Solution 2: Recursive, preorder

    /**
     * Recursive O(n); O(h)
     * use preorder traversal with root and a StringBuilder object.
     * Append current node's val and a delimiter.
     * Then recurse down to left and right subtrees.
     * Base case:
     * If node is null, append a null node and a delimiter.
     * => 1,2,#,#,3,4,#,#,5,#,#,
     */
    public String serializeB(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        return buildString(root, sb).toString();
    }

    private StringBuilder buildString(TreeNode root, StringBuilder sb) {
        if (root == null) {
            return sb.append("#").append(",");
        }

        sb.append(root.val).append(",");
        buildString(root.left, sb).append(",");
        buildString(root.right, sb);
        return sb;
    }

    /**
     * Recursive O(n); O(n)
     * Same as preorder traversal.
     * Split data by "," and use Queue to store strings.
     * Poll a value string from the queue.
     * If null node, return null.
     * Create a tree node with value.
     * Then build left and right subtrees recursively.
     */
    public TreeNode deserializeB(String data) {
        Queue<String> queue = new LinkedList<>(Arrays.asList(data.split(",")));
        return buildTree(queue);
    }

    private TreeNode buildTree(Queue<String> queue) {
        String head = queue.poll();
        if ("#".equals(head)) {
            return null;
        }

        TreeNode root = new TreeNode(Integer.valueOf(head));
        root.left = buildTree(queue);
        root.right = buildTree(queue);
        return root;
    }