606. Construct String from Binary Tree[Easy, DFS, preorder traversal]

weekly test 35的第二题，easy的，但我没做出来；一方面不太理解题目的one-to-one mapping是什么意思（现在理解的是，如果左孩子没有 右孩子有的情况，就保留左孩子的空braces，否则无法区分右孩子），一方面实在怕了递归了。

答案：

public String tree2str(TreeNode t) {
        return preOrder(t);
    }

    private String preOrder(TreeNode t) {
        if (t == null)
            return "";
        //右孩子是空的情况，为了避免加括号，避免preorder右孩子
        if (t.left != null && t.right == null) {
            return t.val + "(" + preOrder(t.left) + ")";
        }
        if (t.left == null && t.right == null) {
            return t.val + "";
        }
        return t.val + "(" + preOrder(t.left) + ")(" + preOrder(t.right) + ")";
    }

或者：
https://discuss.leetcode.com/topic/91311/java-simple-recursion

public String tree2str(TreeNode t) {
        StringBuilder sb = new StringBuilder();
        helper(sb,t);
        return sb.toString();
    }
    public void helper(StringBuilder sb,TreeNode t){
        if(t!=null){
            sb.append(t.val);
            if(t.left!=null||t.right!=null){
                sb.append("(");
                helper(sb,t.left);
                sb.append(")");
                if(t.right!=null){
                    sb.append("(");
                helper(sb,t.right);
                sb.append(")");
                }
            }
        }
    }

或者：

public class Solution {
    public String tree2str(TreeNode t) {
        if (t == null) return "";
        
        String result = t.val + "";
        
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        
        if (left == "" && right == "") return result;
        if (left == "") return result +"()" + "(" + right + ")";
        if (right == "") return result + "(" + left + ")";
        return result + "(" + left + ")" + "(" + right + ")";
    }
}