leetcode -- 606. Construct String from Binary Tree【递归 + 字符串优化】

题目

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

给定一棵树，对其进行先序遍历，将结果表示成字符串的形式。

分析及解答

解法1：(对字符串优化不够，浪费了大量时间)

public class Solution {    public String tree2str(TreeNode t) {       if(t == null) return "";StringBuilder result = new StringBuilder(String.valueOf(t.val));String leftResult = tree2str(t.left);String rightResult = tree2str(t.right);if(leftResult == "" && rightResult == ""){return result.toString();}else{if(leftResult == ""){result.append("()");}else{                result.append("(");result.append(leftResult);                result.append(")");}if(rightResult != ""){result.append("(");result.append(rightResult);result.append(")");}return result.toString();}    }}

解法2：（对java字符串进行优化）

• 思路未变，只是增加了字符串的优化（String 变为了 StringBuilder）
  

/** * 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();        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(")");                }            }        }    }}