python--leetcode606. Construct String from Binary Tree

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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本文介绍了一道LeetCode题目，要求将二叉树以特定规则转换为字符串形式，通过前序遍历的方式实现，同时避免不必要的空括号出现。提供了Python代码示例。

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最近确实忙啊，半个月没上过leetcode了，今天趁代码暂时写完了来刷一刷题。不多说了看题。

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.

这题题目的意思就是说，给你一颗二叉树，让你生成一个字符串，该字符串中父节点后用括号把所有子节点括住，如果有空节点用（）代替，并且不能有多余的（）出现。

解题思路：还是前序遍历二叉树，并且要稍微做一些判断，来防止多余（）出现。

# 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 ""
        res = ""
        left = self.tree2str(t.left)
        right = self.tree2str(t.right)
        if left or right:
            res += "(%s)" % left
        if right:
            res += "(%s)" % right
        return str(t.val) + res