题目描述：请实现一个函数按照之字形打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右至左的顺序打印，第三行按照从左到右的顺序打印，其他行以此类推。

Solutions

类似于分行打印，只不过各一次反过来打印一次：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
def Print(self, pRoot):
# write code here
if pRoot is None:
return []
res = []
queue = [pRoot]
b_left_to_right = 0
while queue:
level = []
size = len(queue)
for i in range(size):
node = queue.pop(0)
level.append(node.val)
if node.left is not None:
queue.append(node.left)
if node.right is not None:
queue.append(node.right)
if b_left_to_right & 1:
level = level[::-1]
b_left_to_right += 1
res.append(level)
return res
# 运行时间：21ms
# 占用内存：5728k


剑指 Offer 的解法，尚未细看：

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
def Print(self, pRoot):
# write code here
if not pRoot:
return []
res=[]
nodes=[pRoot]
right=True

while nodes:
curStack,nextStack=[],[]
if right:
for node in nodes:
curStack.append(node.val)
if node.left:
nextStack.append(node.left)
if node.right:
nextStack.append(node.right)
else:
for node in nodes:
curStack.append(node.val)
if node.right:
nextStack.append(node.right)
if node.left:
nextStack.append(node.left)
res.append(curStack)
nextStack.reverse()
right=not right
nodes=nextStack
return res