The base version was fine but then they asked for O(1) space and I kind of stalled.
Clarify that 'leaf nodes' means nodes with no children, and that we need to set their next pointers to link them in left-to-right order. Use a level-order traversal that processes each level using the next pointers already established for the previous level, achieving O(1) extra space. Alternatively, if the tree is a perfect binary tree, use a recursive approach that connects subtrees, but for general binary trees, the iterative level-order approach is more robust.
Pro tip: Mention that this is a variation of the classic 'Populating Next Right Pointers in Each Node' problem, but with the twist of only connecting leaves. Emphasize that you must handle the case where the tree is not perfect and that you need to skip internal nodes when linking.
Confirm that leaf nodes are nodes with no children, and that next pointers should link leaves in left-to-right order. Ask if the tree is guaranteed to be perfect or if it can be any binary tree, as this affects the approach.
Explain that you will traverse the tree level by level using the next pointers of the current level to move to the next node, and set next pointers for the next level. To connect only leaves, you need to track the first leaf of the next level and the previous leaf to link them.
Maintain a 'prevLeaf' pointer to the last leaf processed and a 'head' pointer to the first leaf of the next level. For each node in the current level, if it has a left child, process it; if it has a right child, process it. When processing a child, if it is a leaf, link it to prevLeaf and update prevLeaf; otherwise, continue. Use the next pointers to move across the current level.
Discuss edge cases: empty tree, single node (which is a leaf), and trees where leaves are at different depths. Confirm that the time complexity is O(n) and space is O(1) since we only use a few pointers.
Walk through a small example, such as a tree with root 1, left child 2 (leaf), right child 3 with left child 4 (leaf) and right child 5 (leaf). Show how the next pointers are set: 2 -> 4 -> 5.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.