題目：

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.
 

解釋： 注意，有右子樹的時候，左子樹一定要有值，就算是空的()也要有，但是隻有左子樹沒有右子樹的時候，只寫出左子樹即可，右子樹的空字串''，基於這樣的原則寫if條件。 python程式碼，用到了format()函式：

# 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 t==None:
            return ''
        left,right='',''
        if t.left or t.right:
            left='({})'.format(self.tree2str(t.left))
        if t.right:
            right='({})'.format(self.tree2str(t.right))
        return  '{}{}{}'.format(t.val,left,right) 

更常規的python程式碼：

# 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 t==None:
            return ''
        left,right='',''
        if t.left or t.right:
            left='('+self.tree2str(t.left)+')'
        if t.right:
            right='('+self.tree2str(t.right)+')'
        return  str(t.val)+left+right

c++程式碼：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    string tree2str(TreeNode* t) {
        if (!t)
            return "";
        string left ="",right="";
        if (t->left||t->right)
            left="("+tree2str(t->left)+")";
        if (t->right)
            right="("+tree2str(t->right)+")";
        return to_string(t->val)+left+right;
    }
};

總結： python的format()函式還是蠻有意思的。