# 103. Binary tree zigzag level order traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree [3,9,20,null,null,15,7],

``````3
``````

/
9 20 /
15 7

return its zigzag level order traversal as:

[ [3], [20,9], [15,7] ]

• code
``````def zigzagLevelOrder(self, root):
queue = collections.deque([root])
res = []
while queue:
r = []
for _ in range(len(queue)):
q = queue.popleft()
if q:
r.append(q.val)
queue.append(q.left)
queue.append(q.right)
r = r[::-1] if len(res) % 2 else r
if r:
res.append(r)
return res
``````
• code
``````class Solution:
def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
direction = 1
res = []
cur_level = [root]
while cur_level:
cur_level_value = [node.val for node in cur_level]
cur_level = [i for node in cur_level for i in [node.left, node.right] if i]
if direction == 1:
res.append(cur_level_value)
direction = 0
elif direction == 0:
res.append(reversed(cur_level_value))
direction = 1
return res

``````
• c
``````class Solution:
def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []

def traversal(root, level, res):
if not root:
return
if len(res) <= level:
res.append([])

if level%2 == 0:
res[level].append(root.val)
# flag = 1
elif level %2 == 1:
res[level].insert(0, root.val)

traversal(root.left, level+1, res)
traversal(root.right, level+1, res)

res = [[]]
traversal(root, 0, res)
return res
``````