Invert binary tree in Python

InvertBinaryTree.py

from dataclasses import dataclass
from typing import Any
from xml.etree.ElementTree import Element


@dataclass
class Node():
    val: Any
    right: "Node"
    left: "Node"

    def __str__(self):  # DFS recursive-print
        return " ".join((str(self.left), str(self.val), str(self.right))).replace("None", "")

"""
      1
     / \
    2   3
   / \ / \
  4  5 6  7
"""

# build tree
tree = Node(1, None, None)
tree.left = Node(2, None, None)
tree.right = Node(3, None, None)
tree.left.left = Node(4, None, None)
tree.left.right = Node(5, None, None)
tree.right.left = Node(6, None, None)
tree.right.right = Node(7, None, None)

print(tree)

# Option 1 (sensible): recursively swap left and right properties

def invert(root: Node):
    if not root:
        return
    root.left, root.right = invert(root.right), invert(root.left)
    return root

# Option 2 The Enterprise way: 

def enterpriseInvert(root: Node):
    import xml.etree.ElementTree as ET
    def btreeToElementTree(root: Node, el: Element):
        el.attrib['val'] = str(root.val)
        if root.left:
            left = ET.SubElement(el, 'left')
            btreeToElementTree(root.left, left)
        if root.right:
            right = ET.SubElement(el, 'right')
            btreeToElementTree(root.right, right)
        return el
    
    tree = btreeToElementTree(root, ET.Element('tree'))

    # Swap left and right via XPath searches
    lefts = tree.findall(".//left")
    rights = tree.findall(".//right")
    for left in lefts:
        left.tag = "right"
    for right in rights:
        right.tag = "left"

    return ET.tostring(tree)


print(invert(tree))
print(enterpriseInvert(tree))

"""
Prints

 4  2  5  1  6  3  7 
 7  3  6  1  5  2  4 
b'<tree val="1"><right val="3"><right val="7" /><left val="6" /></right><left val="2"><right val="5" /><left val="4" /></left></tree>'
"""