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.
根據(jù)給定的二叉樹返回根據(jù)其建立的字符串，建立原則按照先序序列順序，有子節(jié)點的用括號括起來，葉子節(jié)點的孩子括號省略。特別的，若左孩子不存在而有孩子存在，則左孩子的括號不可省略。

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
遞歸調(diào)用即可。

/**
 * 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 tree2str(TreeNode* t) {
        if(t==nullptr) return "";
        string res;
        res+=to_string(t->val);
        if(t->left) res+="("+tree2str(t->left)+")";
        else if( t->right) res+="()";
        if(t->right) res+="("+tree2str(t->right)+")";
        return res;
    }
};