[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