We can solve the problem easily based on last problem: just reverse the odd level. We can reverse it in process or in result.
Solution:
# T:O(n) S:O(n)
class Solution:
# @param root, a tree node
# @return a list of lists of integers
def preorder(self, root, level, res):
if root:
if len(res) < level+1: res.append([])
if level % 2 != 0:
res[level].insert(0, root.val)
else:
res[level].append(root.val)
self.preorder(root.left, level+1, res)
self.preorder(root.right, level+1, res)
def zigzagLevelOrder(self, root):
res=[]
self.preorder(root, 0, res)
return res
Run Time: 76 ms
No comments:
Post a Comment