给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)
实现思路:
先实现普通的二叉树层次遍历,然后将结果逆序。
# 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 levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
queue = [root] # 放进去的是节点对象
res = []
if not root:
return []
while queue:
templist = [] # 用来存放每一层的值
templen = len(queue)
# 对于每一层的 每一个值,每次都找出其左右节点,放进队列
for i in range(templen):
temp = queue.pop(0)
templist.append(temp.val)
if temp.left:
queue.append(temp.left)
if temp.right:
queue.append(temp.right)
res.append(templist)
result = []
for i in range(len(res)-1,-1,-1):#将普通层次遍历的结果逆序
result.append(res[i])
return result
n1 = TreeNode(4)
n2 = TreeNode(2)
n3 = TreeNode(7)
n4 = TreeNode(1)
n5 = TreeNode(3)
n1.left = n2
n1.right = n3
n2.left = n4
n2.right = n5
s = Solution()
result = s.levelOrderBottom(n1)
print(result)