[LeetCode 297] Serialize and Deserialize Binary Tree

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 as how 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. preorder traversal, use another helper class(StringToken) to split string, and record current position

2. level order traversal

public class Codec {

        
 // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        helptoserialize(root, sb);
        return sb.toString();
    }

    public void helptoserialize(TreeNode root, StringBuilder sb) {
        if(root == null) {
            sb.append("# ");
        } else {
            sb.append(root.val);
            sb.append(" ");
            helptoserialize(root.left, sb);
            helptoserialize(root.right, sb);
        }
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        TreeNode root = helptodeserialize(new StringToken(data, " "));
        return root;
    }
    
    public TreeNode helptodeserialize(StringToken token) {
        if(!token.hasMoreElement()) return null;
        String t1 = token.nextElement();
        if(t1.equals("#")) return null;
        TreeNode root = new TreeNode(Integer.valueOf(t1));
        root.left = helptodeserialize(token);
        root.right = helptodeserialize(token);
        return root;
    }

    class StringToken {
        public int len = 0;
        ArrayList<String> dataStrings = new ArrayList<>();
        public StringToken (String data, String spliter) {
            String[] datas = data.split(spliter);
            for(int i=0;i<datas.length;i++) {
                dataStrings.add(datas[i]);
            }
        }
        public boolean hasMoreElement() {
            return len<dataStrings.size();
        }
        public String nextElement() {
            return dataStrings.get(len++);
        }
    }
}