【Leetcode】【python】Binary Tree Zigzag Level Order Traversal

题目大意

按之字形遍历二叉树（一正一反）

解题思路

来自：链接
解题思路：这道题和上一题层序遍历那道题差不多，区别只是在于奇数层的节点要翻转过来存入数组。
代码：

代码

BFS

class Solution(object):    def zigzagLevelOrder(self, root):        tree = []        if not root:            return tree        curr_level = [root]        direction = 'L'        # print(type(root), type(curr_level))  # (<class 'precompiled.treenode.TreeNode'>, <type 'list'>)        # print(curr_level)  # 作为list，却并不能遍历整个树        while curr_level:            level_list = []            next_level = []            for temp in curr_level:                level_list.append(temp.val)                if temp.left:                    next_level.append(temp.left)                if temp.right:                    next_level.append(temp.right)            if direction == 'L':                tree.append(level_list)                direction = 'R'            else:                tree.append(level_list[::-1])                direction = 'L'            curr_level = next_level        return tree

DFS

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @return a list of lists of integers    def preorder(self, root, level, res):        if root:            if len(res) < level+1: res.append([])            if level % 2 == 0: res[level].append(root.val)            else: res[level].insert(0, root.val) # 向0位置插入            self.preorder(root.left, level+1, res)            self.preorder(root.right, level+1, res)    def zigzagLevelOrder(self, root):        res=[]        self.preorder(root, 0, res)        return res

总结

1. insert()方法语法：
   list.insert(index, obj)
   参数
   index – 对象 obj 需要插入的索引位置。
   obj – 要插入列表中的对象。
2. tree.append(level_list[::-1])将数组倒序插入