給定一個(gè)二叉樹(shù)，返回其按層次遍歷的節(jié)點(diǎn)值。 （即逐層地，從左到右訪問(wèn)所有節(jié)點(diǎn)）游盲。
例如:
給定二叉樹(shù): [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其層次遍歷結(jié)果：
[
[3],
[9,20],
[15,7]
]
雙循環(huán)，廣度
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
if root == None:
return res
q = [root]
while len(q) != 0:
res.append([node.val for node in q])
new_q = []
for node in q:
if node.left:
new_q.append(node.left)
if node.right:
new_q.append(node.right)
q = new_q
return res
遞歸蛮粮，深度
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
self.dfs(root, 0, res)
return res
def dfs(self, root, depth, res):
if root == None:
return res
if len(res) < depth+1:
res.append([])
res[depth].append(root.val)
self.dfs(root.left, depth+1, res)
self.dfs(root.right, depth+1, res)
