Problem URL: Construct string from binary tree

We will use recursive DFS to traverse the tree and append the result to a string and return it.

# 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:
        left, right="", ""

        if root.left:
            left = f"({self.tree2str(root.left)})"
        if root.right:
            right = f"({self.tree2str(root.right)})"

        if root.left is None and root.right is not None:
            return str(root.val) + "()" + right

        return str(root.val) + left + right

Time Complexity: O(n)
Space Complexity: O(n)