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.

根据二叉树创建字符串。题意是将给的二叉树按照前序遍历的顺序序列化。

这道题既然说了是按照前序遍历的顺序序列化，那么思路也就是前序遍历。我们需要一个stringbuilder和一个helper函数。唯一需要注意的地方是如果某个节点没有左孩子但是有右孩子（39行），需要记得创建一个空的括号对否则没有左孩子这个信息是无法被表示的。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public String tree2str(TreeNode t) {
18         // corner case
19         if (t == null) {
20             return "";
21         }
22 
23         // normal case
24         StringBuilder sb = new StringBuilder();
25         helper(t, sb);
26         return sb.toString();
27     }
28 
29     private void helper(TreeNode root, StringBuilder sb) {
30         sb.append(root.val);
31         if (root.left == null && root.right == null) {
32             return;
33         }
34         if (root.left != null) {
35             sb.append("(");
36             helper(root.left, sb);
37             sb.append(")");
38         }
39         if (root.right != null) {
40             if (root.left == null) {
41                 sb.append("()");
42             }
43             sb.append("(");
44             helper(root.right, sb);
45             sb.append(")");
46         }
47     }
48 }

 

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