0100 - 0199
0103 - Binary Tree Zigzag Level Order Traversal (Medium)
Problem Link
https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
Problem Statement
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
Approach 1: BFS
# 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:
# Level traversal -> BFS
# reverse the list for odd-index level
def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
# if there is no root, then return []
if not root: return []
# init ans and queue with initial node `root`
ans, q = [], [root]
# BFS
while q:
# direction - 1 for even-index level and -1 for odd-index level
d = -1 if len(ans) % 2 == 1 else 1
# put all node values to a list with the correct direction
# and add to `ans`
ans.append([n.val for n in q][::d])
# for each node in the queue,
# we add the left or right node to the queue if applicable
q = [n for node in q for n in (node.left, node.right) if n]
return ans