# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def __init__(self):
self.res = []
self.d = True
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if root == None:
return self.res
self.func(root,0)
for i,each in enumerate(self.res):
if i % 2 != 0:
self.res[i] = self.res[i][::-1]
return self.res
def func(self,root,depth):
if root == None:
return 0
elif depth >= len(self.res):# and depth % 2 == 0:
self.res.append([root.val])
elif depth <len(self.res):
self.res[depth].append(root.val)
self.func(root.left,depth + 1)
self.func(root.right,depth + 1)
Leetcode刷题记录——剑指 Offer 32 - III. 从上到下打印二叉树 III
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转载自blog.csdn.net/weixin_41545780/article/details/107552109
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