Binary Tree Level Order Traversal II
Given a binary tree, return thebottom-up level ordertraversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree[3,9,20,null,null,15,7]
,
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
从底部层序遍历其实还是从顶部开始遍历,只不过存储的方式有所改变,把level每次加到result的最前边:
result.insert(0,[i.val for i in level])
class Solution(object):
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
result =[]
level = [root]
while level:
newlevel = []
result.insert(0,[i.val for i in level])
for n in level:
if n.left:
newlevel.append(n.left)
if n.right:
newlevel.append(n.right)
level = newlevel
return result
或者最后返回result的reverse
return result[::-1]
return list(reversed(result))
Note thatreversed(...)
does not return a list. You can get a reversed list usinglist(reversed(array))
.
class Solution(object):
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
result =[]
level = [root]
while level:
newlevel = []
result.append([i.val for i in level])
for n in level:
if n.left:
newlevel.append(n.left)
if n.right:
newlevel.append(n.right)
level = newlevel
return result[::-1]
Last updated
Was this helpful?