The stack-only constraint is what tripped me up first.
Use two stacks to simulate level-order traversal: one stack for the current level and another for the next level. Process nodes from the current stack, and for each node, push its right child then left child onto the next stack to ensure left-to-right order when popped. Collect nodes that satisfy P at each level and return the list of levels.
Pro tip: Emphasize that using two stacks is a common pattern for level-order traversal without queues, and clarify that the order of pushing children (right then left) is crucial to maintain left-to-right processing. Also, mention that this approach can be extended to handle multiple levels by swapping stacks after each level.
Restate the problem to ensure understanding: level-order traversal using only stacks, collecting nodes that satisfy predicate P at each level, and returning a list of levels with nodes in left-to-right order. Confirm that the tree is binary and that P is a function that takes a node and returns a boolean.
Explain that you will use two stacks: currentLevel and nextLevel. Initialize currentLevel with the root. While currentLevel is not empty, pop nodes from currentLevel, and if they satisfy P, add them to the current level's result list. For each popped node, push its right child then left child onto nextLevel (to ensure left-to-right order when popped). After processing all nodes in currentLevel, swap currentLevel and nextLevel, and add the current level's result to the overall list.
Trace the algorithm on a small binary tree (e.g., root with left and right children, and some grandchildren) to demonstrate how nodes are processed level by level and how the order is maintained. Show how the stacks evolve and how the predicate filters nodes.
Prove that the algorithm processes nodes level by level in left-to-right order. Use induction on levels: assume that when processing level k, the currentLevel stack contains all nodes of level k in left-to-right order (top of stack is leftmost). Then, by pushing right then left children, the nextLevel stack will contain nodes of level k+1 in left-to-right order (top is leftmost). Thus, when swapped, the next iteration processes level k+1 correctly.
Time complexity: O(N), where N is the number of nodes, since each node is pushed and popped exactly once. Space complexity: O(W), where W is the maximum width of the tree, as the stacks hold at most the number of nodes in two consecutive levels. In the worst case (e.g., perfect binary tree), W = O(N), so space is O(N).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Actually liked this follow-up more than the original.
Use two stacks to simulate level-order traversal without queues: one stack for the current level and another for the next level. Alternate the order in which children are pushed (left-to-right or right-to-left) based on the current level's direction to achieve zigzag output.
Pro tip: Clarify that this approach uses O(n) space in the worst case (for a full level), which is optimal for level-order traversal, and mention that it avoids the overhead of queue operations while elegantly handling the zigzag pattern.
Create two stacks: currentLevel and nextLevel. Push the root onto currentLevel. Set a boolean leftToRight to true, indicating the direction for the current level.
While currentLevel is not empty, pop nodes one by one. For each node, output its value. Depending on leftToRight, push its children onto nextLevel in the appropriate order (left then right if leftToRight is true, else right then left).
After processing all nodes in currentLevel, swap currentLevel and nextLevel (i.e., currentLevel = nextLevel, nextLevel = empty stack). Toggle leftToRight to reverse the direction for the next level.
Continue the loop until both stacks are empty, ensuring all levels are traversed and output in zigzag order.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.