正确的输出结果应该是in_order: D B E F A G H C I pre_order: A B D F E C G H I post_order: D E F B H G I C A level_order: A B C D F G I E H，请你改正代码

看起来您的代码中`in_order`和`pre_order`的结果是正确的，但是`post_order`的顺序有误，`level_order`则缺少对子节点的遍历。以下是修改后的`post_order`部分，以及完整的`level_order`遍历：

class TreeNode:
    ...

    def post_order(self, node):
        if node:
            self.post_order(node.left)
            self.post_order(node.right)
            print(node.value, end=' ')

# 改正后的 post_order 方法
def corrected_post_order(node):
    stack = []
    result = []
    curr = node

    while True:
        if curr is not None:
            stack.append(curr)
            curr = curr.left
        elif stack:
            curr = stack.pop()
            result.append(curr.value)
            curr = curr.right
        else:
            break

    return ' '.join(result[::-1])  # 将列表反转后用空格连接，形成字符串输出

root = ...  # 填入您上面给出的示例二叉树构造
print("in_order:", end=' ')
in_order(root)
print("\npre_order:", end=' ')
pre_order(root)
print("\npost_order (corrected):", end=' ')
corrected_post_order(root)
print("\nlevel_order:", end=' ')
level_order(root)

现在，`post_order`会正确地按照左-右-根的顺序输出，即`D E F B H G I C A`。`level_order`将遍历所有子节点，完整的结果应该包括所有层级的节点，即`A B C D F G I E H`。