606. Construct String from Binary Tree 从二叉树中构建字符串

［抄题］：

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.

[暴力解法]：

时间分析：

空间分析：

［奇葩输出条件］：

数字毕竟不熟字符串，数字+双引号""才是，这是计算机思维

［奇葩corner case］：

［思维问题］：

DFS中是可以分为step1 step2的，因为下一次DFS之前肯定会走step2

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

1. 字符串的题判断空的条件是 == ""，头一次见 注意下
2. 先dfs, 后操作，所以没理解到位的事 都是对dfs的结果left right操作

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

1. 检查函数返回值，不是void型就要返回正常值（所以要提前注释）

［总结］：

字符串变量反而不需要加引号 因为已经规范化了，字符串要加

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［英文数据结构或算法，为什么不用别的数据结构或算法］：

stringbuilder不好用，直接用+建字符串就行

［关键模板化代码］：

新建字符串模板

if (left == "" && right == "") return ans;
        //left is null
        if (left == "") return ans + "()" + "(" + right + ")";
        //right is null
        if (right == "") return ans + "(" + left + ")";
        //return
        return ans + "(" + left + ")" + "(" + right + ")";

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

536. Construct Binary Tree from String 反过来：高级版string文字游戏，唉

[代码风格] ：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public String tree2str(TreeNode t) {
        //corner case
        if (t == null) {
            return "";
        }
        //ini
        String ans = t.val + "";
        //dfs
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        //build string
        //exit
        if (left == "" && right == "") return ans;
        //left is null
        if (left == "") return ans + "()" + "(" + right + ")";
        //right is null
        if (right == "") return ans + "(" + left + ")";
        //return
        return ans + "(" + left + ")" + "(" + right + ")";
    }
}