# [Solved] Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).

## Question

Given the `root` of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).

Example 1:

```Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
```

Example 2:

```Input: root = [1]
Output: [[1]]
```

Example 3:

```Input: root = []
Output: []
```

Constraints:

• The number of nodes in the tree is in the range `[0, 2000]`.
• `-100 <= Node.val <= 100`

## Python Solution

```# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
if root==None:return None
res = []
q = [root]
level = 1

while q:
tempq = []
lev = []
while q:
curr = q.pop(0)
lev.append(curr.val)
if curr.left :tempq.append(curr.left)
if curr.right :tempq.append(curr.right)
q = tempq
if not level%2:
res.append(reversed(lev))
else:
res.append(lev)
level+=1
return res```