python_lintcode_69二叉树的层次遍历

69二叉树的层次遍历

题目

给出一棵二叉树，返回其节点值的层次遍历（逐层从左往右访问）

您在真实的面试中是否遇到过这个题？ Yes

样例

给一棵二叉树 {3,9,20,#,#,15,7} ：

.
3
/ \
9  20
/  \
15   7

返回他的分层遍历结果：

[
[3],
[9,20],
[15,7]
]

思路

主要是实现先进先出FIFO

代码

"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""

class Solution:
"""
@param: root: A Tree
@return: Level order a list of lists of integer
"""
def levelOrder(self, root):
# write your code here
if root is None:return []
stack = [root]
yy=[]
#none就出
while stack:
y=[]
#出栈
#每一层
for i in range(len(stack)):
current = stack.pop(0)
y.append(current.val)
if current.left:
#入栈
stack.append(current.left)
if current.right:
stack.append(current.right)
yy.append(y)
return yy

另外，我同学用的队列

"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""

import Queue

class Solution:
"""
@param: root: A Tree
@return: Level order a list of lists of integer
"""
def levelOrder(self, root):
res = []
q = Queue.Queue()
if root:
q.put(root)

while not q.empty():
level = []
length = q.qsize()
for i in range(length):
node = q.get()
level.append(node.val)
if node.left:
q.put(node.left)
if node.right:
q.put(node.right)
res.append(level)
return res