Serialize and Deserialize Binary Tree & BST

摘要：思路理论上说所有遍历的方法都可以。但是为了使和的过程都尽量最简单，是不错的选择。用作为分隔符，来表示。复杂度代码思路这道题和之前不同，一般的树变成了，而且要求是。还是可以用，还是需要分隔符，但是就不需要保存了。

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.

理论上说所有遍历的方法都可以。但是为了使serialize和deserialize的过程都尽量最简单，preorder是不错的选择。serialize的话，dfs比较好写，deserialize的话preorder和bfs比较好写。用“,”作为分隔符，“#”来表示null。例子里serialize之后结果就变成"1,2,3,#,#,4,5"。deserialize的时候用一个queue来保存string。

Time: O(N), Space: O(N)

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        // base case
        if(root == null) return "";
        StringBuilder encoded = new StringBuilder();
        encode(root, encoded);
        return encoded.substring(1).toString();
    }
    
    private void encode(TreeNode root, StringBuilder sb) {
        if(root == null) {
            sb.append(",#");
            return;
        }
        sb.append(",").append(root.val);
        encode(root.left, sb);
        encode(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        // base case
        if(data.length() == 0) return null;
        Queue q = new LinkedList(Arrays.asList(data.split(",")));
        return decode(q);
    }
    
    private TreeNode decode(Queue q) {
        if(q.isEmpty()) return null;
        String cur = q.poll();
        if(cur.equals("#")) return null;
        TreeNode root = new TreeNode(Integer.valueOf(cur));
        root.left = decode(q);
        root.right = decode(q);
        return root;
    }

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.

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.

这道题和之前不同，一般的树变成了BST，而且要求是as compact as possible。还是可以用preorder，还是需要分隔符，但是null就不需要保存了。deserialize部分要变得复杂，left的值总是小于root的值，right的值总是大于root的值，根据这个每次recursion的时候把左边的值都放到另一个queue里面，剩下的就是右边的值。

Time: O(N^2), Space: O(N)

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        // base case
        if(root == null) return "";
        StringBuilder encoded = new StringBuilder();
        encode(root, encoded);
        return encoded.substring(1).toString();
    }
    
    private void encode(TreeNode root, StringBuilder sb) {
        if(root == null) return;
        sb.append(",").append(root.val);
        encode(root.left, sb);
        encode(root.right, sb);
    }
    
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        // base case
        if(data.length() == 0) return null;
        Queue q = new LinkedList();
        for(String s : data.split(",")) q.offer(Integer.valueOf(s));
        return decode(q);
    }
    
    private TreeNode decode(Queue q) {
        if(q.isEmpty()) return null;
        int cur = q.poll();
        TreeNode root = new TreeNode(cur);
        Queue left = new LinkedList(); 
        while(!q.isEmpty() && q.peek() < cur) left.offer(q.poll());
        root.left = decode(left);
        root.right = decode(q);
        return root;
    }