103. 二叉树的锯齿形层次遍历

给定一个二叉树，返回其节点值的锯齿形层次遍历。（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）。
例如：
给定二叉树 [3,9,20,null,null,15,7],

image.png

在层次遍历的基础上判断层数是偶数还是奇数，偶数不变，奇数数组反转

方法1--队列迭代--BFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        helper = [root]#把根节点存储在队列
        res = []
        while helper:
            tmp1 = []#存储每层节点
            tmp2 = [] ###存储左右子树节点
            for node in helper:
                tmp1.append(node.val)
                if node.left:
                    tmp2.append(node.left)
                if node.right:
                    tmp2.append(node.right)
            res.append(tmp1)#这里就是数组里面套数组
            helper = tmp2
        returnResult=[]
        for i, v in enumerate(res):
            if i%2==0:
                returnResult.append(v)
            else:
                returnResult.append(v[::-1])#这里每一个里面是一个数组
                #result=[[1],[2,3],[4,5,6,7]]
        return returnResult

方法2--递归--DFS深层遍历的递归方法

class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
        res = []
        
        def helper(root, depth):
            if not root: return 
            if len(res) == depth:
                res.append([])
            if depth % 2 == 0:res[depth].append(root.val)
            else: res[depth].insert(0, root.val)#倒序打印，用insert一直往前插入
            helper(root.left, depth + 1)
            helper(root.right, depth + 1)
        helper(root, 0)
        return res