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.
前序遍历形成对应字符串：根（左）（右），当右是空的时候，括号可以省略。
递归过去，依次补括号就可以。因为右是空的时候需要省略括号，所以要加 if (t->right) ，左就不需要了。
class Solution { private: string s; public: void PreOrder(TreeNode* t) { if(t) { s += to_string(t->val); if (t->left || t->right) { s += "("; PreOrder(t->left); s += ")"; if (t->right) { s += "("; PreOrder(t->right); s += ")"; } } } } string tree2str(TreeNode* t) { PreOrder(t); return s; } };
