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.

分4種情況考慮：

1. 左右子節點都沒有，不加()

2. 左右子節點都有，不加()

3. 只有左子節點，不加()

4. 只有右子節點，加()

time: O(n), space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
 
public String tree2str(TreeNode t) {
        if(t == null) {
            return "";
        }
        if(t.left == null && t.right == null) {
            return Integer.toString(t.val);
        }
        if(t.right == null) {
            return t.val + "(" + tree2str(t.left) + ")" + tree2str(t.right);
        }
        return t.val + "(" + tree2str(t.left) + ")" + "(" + tree2str(t.right) + ")";
    }
}