Problem: Serialize and Deserialize BST

Posted by Marcy on February 18, 2015

Question

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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

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

Solution

TODO

Code

Here is a sample solution.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        List<String> list = new ArrayList<>();
        serialize(root, list);
        return String.join(" ", list);
    }

    private void serialize(TreeNode root, List<String> list) {
        if(root == null) return;
        list.add(String.valueOf(root.val));
        serialize(root.left, list);
        serialize(root.right, list);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data.length() == 0) return null;
        Queue<Integer> q = new LinkedList<>();
        String[] list = data.split(" ");
        for(String s: list) {
            q.offer(Integer.valueOf(s));
        }

        return deserialize(q, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public TreeNode deserialize(Queue<Integer> q, long min, long max) {
        if(q.isEmpty()) return null;
        if(q.peek() < min || q.peek() > max) return null;
        int val = q.poll();

        TreeNode n = new TreeNode(val);
        n.left = deserialize(q, min, val);
        n.right = deserialize(q, val, max);

        return n;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

Performance

Both serialize and deserialize methods traverse over each node in the tree once. So the time complexity is O(n), and the space complexity is O(n) for using recursion and the queue.