要求
给定一个二叉树,返回其节点值自底向上的层序遍历。 (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)
例如: 给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其自底向上的层序遍历为:
[
[15,7],
[9,20],
[3]
]
代码详解
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
result = list(list())
nodes = [root]
self.generate(nodes,result)
return result[::-1]
def generate(self,nodes,result):
current_layer_values = []
next_layer_node = []
for node in nodes:
if node.left:
next_layer_node.append(node.left)
if node.right:
next_layer_node.append(node.right)
current_layer_values.append(node.val)
result.append(current_layer_values)
if len(next_layer_node) == 0:
return
self.generate(next_layer_node,result)
解题思路:这个题其实就是102题的思路,最后我们将102的输出进行反转即可得到这个题的输出,102题的链接