Very Simple

• Just traverse using BFS and if level is even then store in normal order else store level elements in reversed order

``````# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
# First find out height
height = self.height(root)
result = []
# Here the approch is BFS
# Hence to simulate these conditions we will use simple for loop
for i in range(height):
output = self.printLevelOrder(root,i, [])
if i % 2 == 0:
result.append(output)
else:
result.append(list(reversed(output)))
return result

# This function basically reaches to the level first then appends result to it
def printLevelOrder(self, node, height, result):
if node == None:
return
# If reached to level add val
if(height == 0):
result.append(node.val)
# Else go left and right with one less level
# Here what you are doing is that decreasing height by 1 until you get height 0 which is nothing but desired level
# So lets say we want to reach level 2 then we will pass height as 2 then decrease it till becomes zero and simultanously go deep               into tree at every recursive call
if height > 0:
self.printLevelOrder(node.left, height-1, result)
self.printLevelOrder(node.right, height-1, result)
return result

# Find out height
def height(self, root):
if root == None:
return 0
return 1+ max(self.height(root.left),self.height(root.right))
``````

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