LeetCode 606. Construct String from Binary Tree 二叉树转带括号字符串

LeetCode 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.

题意: 从一颗二叉树转为带括号的字符串。这题是LeetCode 536的姊妹题型，该题目的解法在这里LeetCode 536解法。

解题思路：和536同样，这题的括号的位置，字符串的结构为root.val(root.left.val)(root.right.val)，当left为空时，须要多加一个(), 咱们循环DFS调用function, 先获得当前的node的value，再获得左子树的字符串，和右子树的字符串，用StringBuilder连接起来，用""来判断是否为空，其中值得注意的是- StringBuilder在初始化一个int值时，须要额外添+""，使得它为一个字符串，而不会解析成capacity。

StringBuilder(int initCapacity)
Creates an empty string builder with the specified initial capacity.

public String tree2str(TreeNode t) {
        if (t == null) {
            return "";
        }
        StringBuilder res = new StringBuilder(t.val+"");
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        if (left.equals("") && right.equals("")) {
            return res.toString();
        }
        if (left.equals("") && !right.equals("")) {
            res.append("()(").append(right).append(")");
            return res.toString();
        }
        if (!left.equals("") && right.equals("")) {
            res.append("(").append(left).append(")");
            return res.toString();
        }
        res.append("(").append(left).append(")").append("(").append(right).append(")");
        return res.toString();
        
    }