[LeetCode]Construct String from Binary Tree

内容简介：[LeetCode]Construct String from Binary Tree

题目描述：

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.

题目大意：

将二叉树序列化为字符串，形式为"root(left)(right)"，空节点表示为"()"。所有不产生歧义的空括号可以省去。

解题思路：

递归（Recursion）

Python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def tree2str(self, t):
        """
        :type t: TreeNode
        :rtype: str
        """
        if not t: return ''
        ans = str(t.val)
        if t.left or t.right: ans += '(' + self.tree2str(t.left) + ')'
        if t.right: ans += '(' + self.tree2str(t.right) + ')'
        return ans

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