序列化二叉树

很难缠的一道题，很多解答都没有按照要求序列化字符串。

1. 把二叉树转化为字符串，这步不难，主要是把NULL节点转化为#，并且在所有节点值后面都添加！。
2. 用两个vector，分别存储这一次要添加子节点的节点，和下一步要添加子节点的节点。
3. 在读取的时候，制定了两个函数，一个函数判断字符串第一个数是不是#，第二个函数读取数字，并把指针移向下一个函数。
4. 两个vector轮流一层一层的添加节点，知道最后两个vector都为空。

   class Solution {
   public:
     char* Serialize(TreeNode* root) {
         string serial;
         char* out;
         if (root == 0)
         {
             out = new char[3]();
             out[0] = '#';
             out[1] = '!';
             return out;
         }
         queue queue_of_node;
         queue_of_node.push(root);
         while (!queue_of_node.empty())
         {
             TreeNode* node = queue_of_node.front();
             queue_of_node.pop();
             if (node == 0)
                 serial = serial + '#' + '!';
             else
             {
                 serial = serial + to_string(node->val) + '!';
                 queue_of_node.push(node->left);
                 queue_of_node.push(node->right);
             }
         }
         out = new char[serial.size() + 1];
         memcpy(out, serial.c_str(), serial.size());
         return out;
     }
     TreeNode* Deserialize(char* str) {
         if (str == NULL || str[0] == 0 || str[0] == '#')
             return 0;
         vector vector1;
         vector vector2;
         int index = 0;
         TreeNode* root = new TreeNode(readint_of_str(str, index));
         vector1.push_back(root);
         while (!vector1.empty() || !vector2.empty())
         {
             for (auto i : vector1)
             {
                 addnode(i, "left", str, index, vector2);
                 addnode(i, "right", str, index, vector2);
             }
             vector1.clear();
             for (auto i : vector2)
             {
                 addnode(i, "left", str, index, vector1);
                 addnode(i, "right", str, index, vector1);
             }
             vector2.clear();
         }
         return root;
     }
     void addnode(TreeNode* root, string left_or_right, char* str, int& index, vector& another)
     {
         if (str[index] == 0)
             return;
         if (getchar_of_str(str, index) == '#')
         {
             index = index + 2;
             return;
         }
         else
         {
             TreeNode* temp = new TreeNode(readint_of_str(str, index));
             if (left_or_right == "left")
                 root->left = temp;
             else
                 root->right = temp;
             another.push_back(temp);
         }
     }
     char getchar_of_str(char* str, int& index)
     {
         return str[index];
     }
     int readint_of_str(char* str, int& index)
     {
         int num = 0;
         while (str[index] != '!')
         {
             num = num * 10 + str[index] - '0';
             index++;
         }
         index++;
         return num;
     }
   };