Serialize and Deserialize Binary Tree

题目地址：
https://leetcode.com/problems/serialize-and-deserialize-binary-tree/#/description

题目：

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.

解题思路：
序列化的时候用bfs就好；反序列化的时候也是bfs，但是需要注意的是反序列化的时候每一个node都会对应两个节点，不管是null还是非null，所以存一个boolean的临时变量left，然后每次都flip一下，当queue里面有node并且left是true的时候，就poll出来一个node赋给curr。link: https://discuss.leetcode.com/topic/31264/short-and-straight-forward-bfs-java-code-with-a-queue/6

代码：

// Encodes a tree to a single string.public String serialize(TreeNode root) {
    if(root == null){
        return "";
    }
    Queue<TreeNode> queue = new LinkedList<>();
    TreeNode curr = root;
    queue.offer(curr);
    StringBuilder sb = new StringBuilder();
    while(!queue.isEmpty()){
        curr = queue.poll();
        if(curr == null){
            sb.append("n,");
            continue;
        }
        sb.append(curr.val + ",");
        queue.offer(curr.left);
        queue.offer(curr.right);
    }
    return sb.toString();
}

// Decodes your encoded data to tree.public TreeNode deserialize(String data) {
    if(data == null || data.length() == 0 ){
        return null;
    }
    String[] nodes=data.split(",");
    boolean left=true;
    if(nodes.length == 0 || nodes[0].equals("n")) {
        return null;
    }
    TreeNode root=new TreeNode(Integer.valueOf(nodes[0]));
    TreeNode curr=root;
    Queue<TreeNode> queue = new LinkedList<>();
    for(int i = 1; i <= nodes.length - 1; i++){
        if(!nodes[i].equals("n")){
            TreeNode temp = new TreeNode(Integer.valueOf(nodes[i]));
            if(left){
                curr.left = temp;
            }
            else{
                curr.right = temp;
            }
            queue.offer(temp);
        }
        left = !left;
        if(left && !queue.isEmpty()){
            curr = queue.poll();
        }
    }
    return root;
}