# Construct String From Binary Tree Problem

## Description

LeetCode Problem 606.

Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.

Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:

``````1
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Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, 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:

``````1
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Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
``````

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• -1000 <= Node.val <= 1000

## Sample C++ Code

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