剑指offer-二叉树的序列化和反序列化（困惑）-Java

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今天在牛客上做这个题，发现一个很奇怪的事，以下第一个代码能AC，第二个却不能，但其实我个人觉得第二个更好，也不知道牛客的测试用例是怎么写的。

题目描述

请实现两个函数，分别用来序列化和反序列化二叉树

代码一：

/*public class TreeNode {    int val = 0;    TreeNode left = null;    TreeNode right = null;    public TreeNode(int val) {        this.val = val;    }}*/public class Solution {     public int index = -1;    String Serialize(TreeNode root) {        StringBuffer sb = new StringBuffer();        if(root == null){            sb.append("#,");            return sb.toString();        }        sb.append(root.val + ",");        sb.append(Serialize(root.left));        sb.append(Serialize(root.right));        return sb.toString();  }    TreeNode Deserialize(String str) {        String[] strr = str.split(",");        int len = strr.length;        index++;        if(index >= len){            return null;        }        TreeNode node = null;        if(!strr[index].equals("#")){            node = new TreeNode(Integer.valueOf(strr[index]));            node.left = Deserialize(str);            node.right = Deserialize(str);        }                 return node;  }}

代码二：

/*public class TreeNode {    int val = 0;    TreeNode left = null;    TreeNode right = null;    public TreeNode(int val) {        this.val = val;    }}*/import java.util.*;public class Solution {    String Serialize(TreeNode root) {        if(root == null)return "#,";StringBuffer st = new StringBuffer();st.append(root.val+",");st.append(Serialize(root.left));st.append(Serialize(root.right));return st.toString();  }    TreeNode Deserialize(String str) {       if(str == null || str.length() == 0)return null;Queue<String> qu = new LinkedList<>();String[] arr = str.split(",");int len = arr.length;for(int i = 0; i < len; i++){qu.add(arr[i]);}return buildTree(qu);  }    TreeNode buildTree(Queue<String> qu){if(qu.size() == 0)return null;String cur = qu.poll();if(cur == "#")return null;TreeNode root = new TreeNode(Integer.valueOf(cur));root.left = buildTree(qu);root.right = buildTree(qu);return root;}}

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