剑指offer-二叉树的序列化和反序列化（困惑）-Java

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;
}
}