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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:

img

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]
  • -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