【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 = None

class 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

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。

tree.append(level_list[::-1])将数组倒序插入