解题思路:
序列化没什么好讲的就是前序遍历
然后反序列化生成树类似于树的复制。从父节点到子节点递归。
知道遇到‘#’,返回空

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    string s;
    int j=0;
    void Srtial(TreeNode *root)
    {
        if(root==NULL)
        {
            s+="#!";
            return ;
        }
        s+=to_string(root->val);
        s+="!";
        Srtial(root->left);
        Srtial(root->right);
    }
    char* Serialize(TreeNode *root) {    
        Srtial(root);
        return (char *)s.data();
    }
    TreeNode* Deserialize(char *str) {
        s=str;
        return Deserial();
    }
    TreeNode* Deserial()
    {
        if(s.size()==0)
            return NULL;
        if(s[j]=='!') {
            j++;
            if(j>=s.size())
                return NULL;    
        }
        if(s[j]=='#'){
            j++;
            return NULL;
        }
        int num=0;
        while(s[j]>='0' && s[j]<='9'){
        num=num*10+s[j++]-'0';}
        TreeNode *root=new TreeNode(num);
        root->left=Deserial();
        root->right=Deserial();
        return root;
    }
};