Binary Tree Zigzag Level Order Traversal

To solve the Zigzag Level Order Traversal problem efficiently, I would start by checking if the root is null; if so, return an empty list immediately. For the core logic, I prefer using two stacks to manage the direction changes explicitly, as it avoids the overhead of reversing lists after traversal. I initialize a primary stack with the root node and a secondary empty stack. I also maintain a boolean flag, say 'leftToRight', initialized to true. While the primary stack is not empty, I pop nodes and add their values to a temporary level list. If 'leftToRight' is true, I push the left child first, then the right child onto the secondary stack. Conversely, if 'leftToRight' is false, I push the right child first, then the left. After processing all nodes at the current level, I append the level list to my result, toggle the 'leftToRight' flag, and swap the roles of the two stacks. This ensures the next level processes in the opposite direction. For example, given a tree with root 3, left child 9, and right child 20, the first level is [3]. The second level becomes [20, 9] because we pushed 9 then 20, but popped them in reverse order due to the stack LIFO nature combined with our push order. Finally, I would analyze the complexity: since every node is visited exactly once, the time complexity is O(N), and the space complexity is O(N) to store the result and the maximum width of the tree in the stacks, which aligns well with Google's focus on scalable, efficient algorithms.