You need to construct a binary tree from a string consisting of parenthesis and integers.

The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root’s value and a pair of parenthesis contains a child binary tree with the same structure.

You always start to construct the left child node of the parent first if it exists.

Example:
Input: “4(2(3)(1))(6(5))”
Output: return the tree root node representing the following tree:
4
/
2 6
/ \ /
3 1 5
Note:
There will only be ‘(‘, ‘)’, ‘-‘ and ‘0’ ~ ‘9’ in the input string.
An empty tree is represented by “” instead of “()”.

** 解题思路**
My solution using recursion:
For example, we have string “4(2(3)(1))(6(5))”, to construct a binary tree, we can split the string to 3 parts:
“4”,
“(2(3)(1))”
“(6(5))”

4 is the root val;
“(2(3)(1))” is left tree;
“(6(5))” is right tree;

public TreeNode str2tree(String s) {
      if (s == null || s.length() == 0) return null;
    
      int firstParen = s.indexOf("(");
      int val = firstParen == -1 ?  Integer.parseInt(s) : Integer.parseInt(s.substring(0, firstParen));
      TreeNode cur = new TreeNode(val);
        
      if (firstParen == -1) return cur;
      
      int start = firstParen, leftParenCount = 0;
      for (int i = start; i < s.length(); i++) {
          if (s.charAt(i) == '(') leftParenCount++;
          else if (s.charAt(i) == ')') leftParenCount--;
          
          if (leftParenCount == 0 && start == firstParen)  {
              cur.left = str2tree(s.substring(start + 1, i)); 
              start = i + 1;
          } else if (leftParenCount == 0) {
              cur.right = str2tree(s.substring(start + 1, i));
          }
      }
      return cur;
    }