题:https://leetcode.com/problems/binary-tree-level-order-traversal/description/
题目
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
思路
本质是个广度遍历,两种方法。
1.为了分层会记录当前层queue的个数,每次循环只访问这些结点,将其子结点插入。
2.用两个queue,来实现交替。
code
############ 方法一 ##############
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
queue = []
res = []
if root!=None:
queue.append(root)
while len(queue) != 0:
queuelen = len(queue)
tmpres = []
for i in range(queuelen):
node = queue.pop(0)
tmpres.append(node.val)
if node.left != None:
queue.append(node.left)
if node.right != None:
queue.append(node.right)
res.append(tmpres)
return res
################# 方法二 ##############
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
queueA = []
queueB = []
res = []
if root!=None:
queueA.append(root)
while len(queueA) != 0 or len(queueB) != 0:
tmpres = []
if len(queueA) != 0:
while len(queueA) != 0:
node = queueA.pop(0)
tmpres.append(node.val)
if node.left != None:
queueB.append(node.left)
if node.right != None:
queueB.append(node.right)
else:
while len(queueB) != 0:
node = queueB.pop(0)
tmpres.append(node.val)
if node.left != None:
queueA.append(node.left)
if node.right != None:
queueA.append(node.right)
res.append(tmpres)
return res