题解 | #求二叉树的层序遍历#

class Solution:
    def levelOrder(self , root: TreeNode) -> List[List[int]]:
        # write code here
        current_layer_nodes = [root] if root else []
        ret = []
        # 当当前层不为空列表的时候，说明还没有到底层
        while current_layer_nodes:
            curr_result = []
            next_layer_nodes = []
            # 遍历本层的所有节点（来自于上一层的存储）
            for n in current_layer_nodes:
                curr_result.append(n.val)
                # 存储下一层的节点
                if n.left:
                    next_layer_nodes.append(n.left)
                if n.right:
                    next_layer_nodes.append(n.right)
            # 将当前层的结果存入返回的列表中
            ret.append(curr_result)
            # 重置当前层为下一层，往下走。
            current_layer_nodes = next_layer_nodes
        return ret