Pyhton 二叉树层级遍历

class TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightclass Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:res = []# 空节点，直接返回if not root:return resque = [root]while que:# 暂存下一层节点的队列next_ceng = []# 暂存当前节点的值队列temp = []for node in que:temp.append(node.val)# 获取当前节点的 左右子节点left = node.leftright = node.rightif left:next_ceng.append(left)if right:next_ceng.append(right)# 把当前层节点对应的值加到结果中res.append(temp)# 当前层队列中节点遍历完成，开始处理下层节点que = next_cengreturn res