leetcode 606. Construct String from Binary Tree

You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.

The null node needs to be represented by empty parenthesis pair "()". And you need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:
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)".
Example 2:
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.

思路：题目的意思就是对树的每颗子树的节点用括号括起来。如果左子树不存在，但右子树还在那么左子树用()代替。如果左子树存在，右子树不存在那么右子树的括号就可以省略。
递归遍历，访问右子树的时候看看左子树是不是为空。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    string ans = "";
    void dfs(TreeNode* t) {
        ans += to_string(t->val);
        int mark = 0;
        if (t->left) {
            ans +='(';
            dfs(t->left);
            ans += ')';
            mark = 1;
        }
        if (t->right) {
            if (mark == 0) ans += "()";
            ans +='(';
            dfs(t->right);
            ans += ')';
        }
    }
    string tree2str(TreeNode* t) {
        if (t) dfs(t);
        return ans;
    }
};