请实现两个函数，分别用来序列化和反序列化二叉树

解题思路

序列化二叉树

递归遍历二叉树的节点，空节点使用#代替，节点之间使用逗号隔开，返回字符串

反序列化二叉树

设置序号index，将字符串根据逗号分割为数组，根据index的值来设置树节点的val，如果节点的值为#，则返回空的树节点。

public class SerializeTree {

    int index = -1;
    /**
     * 分别遍历左节点和右节点，空使用#代替，节点之间，隔开
     *
     * @param root
     * @return
     */
    public String Serialize(TreeNode root) {
        if (root == null) {
            return "#";
        } else {
            return root.val + "," + Serialize(root.left) + "," + Serialize(root.right);
        }
    }
    /**
     * 使用index来设置树节点的val值，递归遍历左节点和右节点，如果值是#则表示是空节点，直接返回
     *
     * @param str
     * @return
     */
    TreeNode Deserialize(String str) {
        String[] s = str.split(",");//将序列化之后的序列用，分隔符转化为数组
        index++;//索引每次加一
        int len = s.length;
        if (index > len) {
            return null;
        }
        TreeNode treeNode = null;
        if (!s[index].equals("#")) {//不是叶子节点 继续走 是叶子节点出递归
            treeNode = new TreeNode(Integer.parseInt(s[index]));
            treeNode.left = Deserialize(str);
            treeNode.right = Deserialize(str);
        }
        return treeNode;
    }

    public static void main(String[] args) {
        TreeNode treeNode1 = new TreeNode(1);
        TreeNode treeNode2 = new TreeNode(2);
        TreeNode treeNode3 = new TreeNode(3);
        TreeNode treeNode4 = new TreeNode(4);
        TreeNode treeNode5 = new TreeNode(5);
        TreeNode treeNode6 = new TreeNode(6);

        treeNode1.left = treeNode2;
        treeNode1.right = treeNode3;

        treeNode2.left = treeNode4;
        treeNode3.left = treeNode5;
        treeNode3.right = treeNode6;

        SerializeTree serializeTree = new SerializeTree();

        String str = serializeTree.Serialize(treeNode1);
        TreeNode treeNode = serializeTree.Deserialize(str);
    }
}