Compute left and right views once

Given the root of a binary search tree, output two lists: the nodes visible from the left side and the nodes visible from the right side, from top to bottom. Implement a single-pass traversal that computes both views concurrently. Discuss iterative vs. recursive strategies, how you track depth and first/last-at-depth nodes, and analyze time/space complexity. Handle empty trees and trees with duplicate keys.

Left and Right Views of a Binary Tree in One Pass

Given the root of a binary tree, return two lists: the left view (the first node seen at each depth, scanning top to bottom) and the right view (the last node seen at each depth). Compute BOTH views in a single traversal.

The tree is provided as a LeetCode-style level-order array level_order, where each element is a node value or null (None) to denote a missing child. Reconstruct the tree, then traverse it once.

Return the answer as a list of two lists: [left_view, right_view].

A single level-order BFS works naturally: at each level the first dequeued node is on the left edge and the last is on the right edge. Empty trees return [[], []]. Duplicate keys are allowed and are emitted by value (a single-node level contributes its value to both views).

Example: for [1, 2, 3, 4, null, null, 5] the left view is [1, 2, 4] and the right view is [1, 3, 5], so the answer is [[1, 2, 4], [1, 3, 5]].

Constraints

• 0 <= number of nodes <= 10^4
• Node values fit in a 32-bit signed integer
• Duplicate node values are allowed
• The tree may be empty (level_order is empty or begins with None)
• Must compute both views in a single traversal of the tree

Hints

1. A level-order BFS gives you the views for free: at each level, the first node you dequeue is the leftmost (left view) and the last is the rightmost (right view).
2. Snapshot the queue size at the start of each level so you know which dequeue index is first (i == 0) and which is last (i == n-1).
3. When the tree is a single chain (only-left or only-right children), every level has exactly one node, so the left and right views are identical.
4. A recursive DFS also works: pass the current depth, and for the left view record the first value reached at each new depth (preorder, left-first); for the right view record the first value reached visiting right-first.