LeetCode 606. Construct String from Binary Tree

Description

https://leetcode.com/problems/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.

Explanation

Traverse the tree and append the values and parenthesis. If the node doesn’t have left and right subtrees, we don’t need to append parenthesis. If the node only has right subtrees, we should append parenthesis as a left subtree.

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def tree2str(self, t: TreeNode) -> str:
        
        results = []
        
        self.traverse(t, results)
        
        return "".join(results)
        
    def traverse(self, t, results):
        if not t:
            return
        
        results.append(str(t.val))
        
        if not t.left and not t.right:
            return
        
        results.append("(")
        self.traverse(t.left, results)        
        results.append(")")

        if t.right:        
            results.append("(")
            self.traverse(t.right, results)
            results.append(")")
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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