LeetCode 606. Construct String from Binary Tree 二叉樹轉帶括號字符串

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.

題意: 從一顆二叉樹轉爲帶括號的字符串。這題是LeetCode 536的姊妹題型，該題目的解法在這裏LeetCode 536解法。

解題思路：和536同樣，這題的括號的位置，字符串的結構爲root.val(root.left.val)(root.right.val)，當left爲空時，須要多加一個(), 咱們循環DFS調用function, 先獲得當前的node的value，再獲得左子樹的字符串，和右子樹的字符串，用StringBuilder連接起來，用""來判斷是否爲空，其中值得注意的是- StringBuilder在初始化一個int值時，須要額外添+""，使得它爲一個字符串，而不會解析成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();
        
    }