Binary Tree Spiral Level Order

Return the values of a binary tree one level at a time in alternating directions.
For this exercise, assume the root level is traversed left to right, the next level right to left, and directions alternate thereafter. Group values by their distance from the root and omit missing children. Preserve node positions even when values repeat. An empty tree returns an empty list.

Function
zigzagLevels(root: TreeNode) → int[][]

Examples
Example 1
root = [3,9,20,null,null,15,7]
return = [[3],[20,9],[15,7]]
The root is left-to-right, the second level is right-to-left, and the third returns to left-to-right.
Example 2
root = [1,2,3,4,null,null,5]
return = [[1],[3,2],[4,5]]
Null links are skipped; the second level is [3,2] and the third is [4,5].
Example 3
root = []
return = []
An empty tree has no levels.

Constraints
For this exercise, assume a finite acyclic binary tree with 0 through 500 nodes.
-10^6 <= node.val <= 10^6.

Run a standard BFS with a queue. At each level, record the queue size, pop exactly that many nodes, and collect their values into a list. Push left child then right child every time, no matter the direction. Keep a boolean flag that flips after each level. When the flag says right-to-left, reverse the level list before appending it, or write into a preallocated array from the back. The common pitfall is changing the order you enqueue children based on direction. That breaks the next level's grouping. Another trap is forgetting the empty tree, which must return an empty list, not [[]]. Check Example 2: level two is [3,2] and level three is [4,5], with null links skipped. Complexity is O(n) time and O(n) space. With 500 nodes max, recursion would also work, but BFS is cleaner. If you freeze live, StealthCoder can supply the loop as a hedge, but you only need about fifteen lines.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.