[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).

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Table of Contents

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:

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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

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