lintcode,二叉树的序列化和反序列化

设计一个算法，并编写代码来序列化和反序列化二叉树。将树写入一个文件被称为“序列化”，读取文件后重建同样的二叉树被称为“反序列化”。

如何反序列化或序列化二叉树是没有限制的，你只需要确保可以将二叉树序列化为一个字符串，并且可以将字符串反序列化为原来的树结构。

样例

给出一个测试数据样例， 二叉树{3,9,20,#,#,15,7}，表示如下的树结构：

3

/ \

9 20

/ \

15 7

我们的数据是进行BFS遍历得到的。当你测试结果wrong answer时，你可以作为输入调试你的代码。

你可以采用其他的方法进行序列化和反序列化。

解题思路：序列化过程类似层次遍历，需要注意的是判断节点是否为空，以及字符串拼接的细节。反序列化过程，用index记录对第几个节点做操作，节点保存在arraylist里面，每次判断是否是操作左孩子，如果是右孩子就index++。

一刷ac

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
class Solution {
/**
* This method will be invoked first, you should design your own algorithm
* to serialize a binary tree which denote by a root node to a string which
* can be easily deserialized by your own "deserialize" method later.
*/
public String serialize(TreeNode root) {
if(root == null) return "{}";
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
String res = "{";
while(!queue.isEmpty()){
int size = queue.size();
for(int i = 0; i < size; i++){
TreeNode tmp = queue.poll();
if(tmp == root) {
res += String.valueOf(tmp.val);
queue.offer(tmp.left);
queue.offer(tmp.right);
}
else{
if(tmp == null) res += ",#";
else{
res += ",";
res += String.valueOf(tmp.val);
queue.offer(tmp.left);
queue.offer(tmp.right);
}
}
}
}
res += "}";
return res;
}

/**
* This method will be invoked second, the argument data is what exactly
* you serialized at method "serialize", that means the data is not given by
* system, it's given by your own serialize method. So the format of data is
* designed by yourself, and deserialize it here as you serialize it in
* "serialize" method.
*/
public TreeNode deserialize(String data) {
if(data.equals("{}")){
return null;
}
String[] strs = data.substring(1, data.length()-1).split(",");
TreeNode root = new TreeNode(Integer.parseInt(strs[0]));
ArrayList<TreeNode> al = new ArrayList<TreeNode>();
al.add(root);
boolean isLeft = true;
int index = 0;
for(int i = 1; i < strs.length; i++){
if(strs[i].equals("#")){
if(!isLeft) index++;
isLeft = !isLeft;
}else{
TreeNode node = al.get(index);
TreeNode tmp = new TreeNode(Integer.parseInt(strs[i]));
if(isLeft) node.left = tmp;
else node.right = tmp;
al.add(tmp);
if(!isLeft) index++;
isLeft = !isLeft;
}
}

return root;
}
}