【LeetCode】Add to List 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.

题目是把二叉树转成带括号的字符串显示。

class Solution {public:    string ans;    void  DFS(TreeNode* t){        if(t==NULL)return;        ans+=to_string(t->val);        if(t->left||t->right){  //如果左子树为空，右子树不为空，左边的括号还是需要打印的            ans+='(';            DFS(t->left);               ans+=')';        }        if(t->right){            ans+='(';            DFS(t->right);            ans+=')';        }    }    string tree2str(TreeNode* t) {        DFS(t);        return ans;    }};