We can use a recursive approach. For each node, we swap its left and right children and then recursively call the function for both.

1. If root is None, return None.
2. Swap root.left and root.right.
3. Recursively call invert_tree(root.left).
4. Recursively call invert_tree(root.right).
5. Return root.

• Time Complexity: O(N).
• Space Complexity: O(H).

def invert_tree(root):
    if not root:
        return None
        
    # Swap children
    root.left, root.right = root.right, root.left
    
    # Recurse
    invert_tree(root.left)
    invert_tree(root.right)
    
    return root