LeetCode: 606. Construct String from Binary Tree

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.

自己的答案，执行的时间是31ms：

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

速度最快的答案，执行时间是13ms：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public String tree2str(TreeNode t) {        StringBuilder sb = new StringBuilder();        dfs(t, sb);        return sb.toString();    }    private void dfs(TreeNode cur, StringBuilder sb) {        if (cur == null) {            return;        }        sb.append(cur.val);        if (cur.left != null) {            sb.append('(');            dfs(cur.left, sb);            sb.append(')');        }        if (cur.right != null) {            if (cur.left == null) {                sb.append("()");            }            sb.append('(');            dfs(cur.right, sb);            sb.append(')');        }    }}

Solution