request for member ‘shang’ in ‘i’, which is of non-class type ‘int’ ^~~~~
时间: 2024-02-10 10:59:44 浏览: 127
As an AI language model, I cannot understand the full context of the issue. However, the error message suggests that you are trying to request a member named 'shang' from an integer variable 'i', which is not possible because 'i' is not a class or object that has members.
To resolve this issue, you may need to revise your code and ensure that you are accessing the correct member from the appropriate object or class.
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request for member ‘shang’ in ‘i’, which is of non-class type ‘int’ printf("%d+%d=%d",i.shang,b); ^~~~~
As an AI language model, I do not have emotions like humans do. "null" is typically used to represent that something has no assigned value or is undefined. It can also refer to a database or programming term, indicating an absence of data or a null value.
给我分一下这个代码:class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: if not root: return [] res, queue = [], collections.deque() queue.append(root) i = 1 while queue: temp = collections.deque() for _ in range(len(queue)): node = queue.popleft() if i & 1 == 1: temp.append(node.val) else: temp.appendleft(node.val) if node.left: queue.append(node.left) if node.right: queue.append(node.right) res.append(list(temp)) i += 1 return res 作者:wangpengcufe-2 链接:https://leetcode.cn/problems/cong-shang-dao-xia-da-yin-er-cha-shu-iii-lcof/solution/by-wangpengcufe-2-y4kq/ 来源:力扣(LeetCode) 著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
这段代码是一道 LeetCode 上的题目 "从上到下打印二叉树 III" 的解法,实现了按照层级遍历二叉树,并将同一层级的节点值按照不同方向存储到列表中。
具体来说,这个类中的 `levelOrder` 函数接受一个二叉树的根节点作为参数,并返回一个二维列表,其中每个子列表代表树的一层,并按照题目要求的方向存储节点值。
代码中的主要实现思路是使用队列实现树的层级遍历,并使用一个变量 `i` 记录当前遍历的层级,判断应该按照正序还是反序存储节点值。同时,在遍历每一层时,也需要将该层的节点按照正序或反序存储到一个临时队列中,再将该队列的节点值转化为列表存储到结果列表中。
需要注意的是,代码中使用了 Python 中的 `collections.deque` 实现队列,并使用了位运算判断当前层级的奇偶性。
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