Input: Binary tree: [1,2,3,4]
       1
     /   \
    2     3
   /    
  4     

Output: "1(2(4))(3)"
  
Explanation: Originallay it needs to be "1(2(4)())(3()())",   
but you need to omit all the unnecessary empty parenthesis pairs.   
And it will be "1(2(4))(3)".

Input: Binary tree: [1,2,3,null,4]
       1
     /   \
    2     3
     \  
      4 

Output: "1(2()(4))(3)"
  
Explanation: Almost the same as the first example,   
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

class Solution {
public:
    string tree2str(TreeNode* t) {
        if (!t) return "";
        string res = "";
        helper(t, res);
        return string(res.begin() + 1, res.end() - 1);
    }
    void helper(TreeNode* t, string& res) {
        if (!t) return;
        res += "(" + to_string(t->val);
        if (!t->left && t->right) res += "()";
        helper(t->left, res);
        helper(t->right, res);
        res += ")";
    }
};

class Solution {
public:
    string tree2str(TreeNode* t) {
        if (!t) return "";
        string res = to_string(t->val);
        string left = tree2str(t->left), right = tree2str(t->right);
        if (left == "" && right == "") return res;
        if (left == "") return res + "()" + "(" + right + ")";
        if (right == "") return res + "(" + left + ")";
        return res + "(" + left + ")" + "(" + right + ")";
    }
};

class Solution {
public:
    string tree2str(TreeNode* t) {
        if (!t) return "";
        string res = to_string(t->val);
        if (!t->left && !t->right) return res;
        res += "(" + tree2str(t->left) + ")";
        if (t->right) res += "(" + tree2str(t->right) + ")";
        return res;
    }
};