给定一棵二叉树，使用非递归的方式实现二叉树左右子树交换，并输出后序遍历结果,要求完善以下代码

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def postorderTraversal(root):
    if not root:
        return []

    stack = []
    output = []
    stack.append(root)

    while stack:
        current_node = stack.pop()
        output.append(current_node.val)

        if current_node.left:
            stack.append(current_node.left)

        if current_node.right:
            stack.append(current_node.right)

    return output[::-1]

def invertTree(root):
    if not root:
        return None

    stack = []
    stack.append(root)

    while stack:
        current_node = stack.pop()
        temp = current_node.left
        current_node.left = current_node.right
        current_node.right = temp

        if current_node.left:
            stack.append(current_node.left)

        if current_node.right:
            stack.append(current_node.right)

    return root

# 测试代码
root = TreeNode(4)
root.left = TreeNode(2)
root.right = TreeNode(7)
root.left.left = TreeNode(1)
root.left.right = TreeNode(3)
root.right.left = TreeNode(6)
root.right.right = TreeNode(9)

print("原二叉树后序遍历结果：", postorderTraversal(root))
invertTree(root)
print("翻转后二叉树后序遍历结果：", postorderTraversal(root))

运行结果：

原二叉树后序遍历结果： [1, 3, 2, 6, 9, 7, 4]
翻转后二叉树后序遍历结果： [9, 6, 7, 3, 1, 2, 4]