103.二叉树的锯齿形层序遍历
bfs, https://leetcode.cn/problems/binary-tree-zigzag-level-order-traversal/
给你二叉树的根节点 root ,返回其节点值的 锯齿形层序遍历 。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
示例 1:

输入:root = [3,9,20,null,null,15,7]
输出:[[3],[20,9],[15,7]]示例 2:
输入:root = [1]
输出:[[1]]示例 3:
输入:root = []
输出:[]提示:
- 树中节点数目在范围
[0, 2000]内 -100 <= Node.val <= 100
python
from collections import deque
from typing import Optional, List
# 定义二叉树节点(LeetCode 会自动提供这个定义)
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return []
result = []
queue = deque([root])
left_to_right = True
while queue:
level_size = len(queue)
level_nodes = deque()
for _ in range(level_size):
node = queue.popleft()
if left_to_right:
level_nodes.append(node.val)
else:
level_nodes.appendleft(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(list(level_nodes))
left_to_right = not left_to_right
return result