Invert Binary Tree

To invert this binary tree, I need to swap the left and right children of every node recursively. Let's start by handling the base case: if the root is null, we simply return null. Otherwise, I'll create a temporary variable to hold the current left child, assign the right child to the left position, and then assign the temporary variable to the right position. After performing this swap, I will recursively call the function on both the new left and right subtrees. For example, if we have a root with value 4, left child 2, and right child 7, after the first step, 4's left becomes 7 and right becomes 2. Then we recurse into those nodes. If we chose an iterative approach instead, we would use a queue to perform a level-order traversal, swapping children as we dequeue each node. This ensures we process every node exactly once. The time complexity for either approach is O(n) because we visit every node in the tree. The space complexity depends on the implementation; recursion uses O(h) space for the call stack, where h is the tree height, while the iterative queue method uses O(w) space, where w is the maximum width of the tree. Given Netflix's focus on scalable and readable systems, I would likely present the recursive solution first for its elegance, but be prepared to discuss the iterative version if deep recursion stacks were a concern in their specific infrastructure context.