606 - Construct String from Binary Tree

Details

Key	Value
Link	https://leetcode.com/problems/construct-string-from-binary-tree/
Language	Python 3
Runtime	65 ms, faster than 77.11% of Python3 online submissions for Construct String from Binary Tree
Memory Usage	16.4 MB, less than 72.23% of Python3 online submissions for Construct String from Binary Tree
Datastructures	TreeNode
Algorithms	DFS w/ Recursion
Complexity	Time: O(N) Memory: O(N)

Procedure

1. ...

Code

Option 1: Faster

# 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, root: Optional[TreeNode]) -> str:
        if not root: return ""
        
        result = str(root.val)
        if root.left:
            result += f'({self.tree2str(root.left)})'
            if root.right:
                result += f'({self.tree2str(root.right)})'
        elif root.right:
            result += f'()({self.tree2str(root.right)})'
        
        return result

Option 2: Terse

# 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, root: Optional[TreeNode]) -> str:
        if root is None: return ''
        if root.left  == None and root.right == None: return str(root.val)        
        if root.right == None: return f'{str(root.val)}({self.tree2str(root.left)})'
        return f'{str(root.val)}({self.tree2str(root.left)})({self.tree2str(root.right)})'