lintcode : 二叉树的序列化和反序列化

题目

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

参考九章程序

看看注释就理解了，但是我表示自己想不出来Java

/**
 * 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) {
        // write your code here
        if( root == null)
            return "{}";
        ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
        queue.add(root);
        // 将二叉树的个节点按照从上到下、从左到有的存储在queue中
        for(int i=0;i<queue.size();i++){
            TreeNode q = queue.get(i);
            if(q== null)
                continue;
            queue.add(q.left);
            queue.add(q.right);
        }
        // 去除叶子节点的左右孩子，这个孩子是空值
        while(queue.get(queue.size() - 1) == null){
            queue.remove(queue.size() - 1);
        }
        // 遍历queue把转换成字符串
        StringBuilder sb = new StringBuilder();
        sb.append("{");
        sb.append(queue.get(0).val);
        for(int i=1;i<queue.size(); i++){
            TreeNode q = queue.get(i);
            if(q!= null){
                sb.append(",");
                sb.append(q.val);
            }else{
                sb.append(",#");
            }
        }
        sb.append("}");
        return sb.toString();
    }
    
    /**
     * 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) {
        // write your code here
        if(data == "{}")
            return null;
        // 以逗号分割
        String[] vals = data.substring(1,data.length()-1).split(",");
        ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
        // 根节点 
        TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
        queue.add(root);
        int index = 0;
        boolean isLeftChild = true;
        for (int i = 1; i < vals.length; i++) {
            if (!vals[i].equals("#")) {
                TreeNode node = new TreeNode(Integer.parseInt(vals[i]));
                if (isLeftChild) {
                    queue.get(index).left = node;
                } else {
                    queue.get(index).right = node;
                }
                queue.add(node);
            }
            if (!isLeftChild) {
                index++;
            }
            isLeftChild = !isLeftChild;
        }
        return root;
    }
}