Populating Next Right Pointers in Each Node - Video Solutions

Problem Overview: Given a perfect binary tree, connect every node's next pointer to its immediate neighbor on the same level. If there is no neighbor, the pointer should be NULL. The tree structure guarantees every parent has two children and all leaves are on the same level.

Approach 1: Level Order Traversal (Breadth-First Search) (Time: O(n), Space: O(n))

This approach processes the tree level by level using a queue. Push the root into the queue, then iterate through nodes level-wise. For each level, track the previous node and set its next pointer to the current node as you dequeue elements. After finishing a level, the last node naturally points to NULL. This method is straightforward and works for any binary tree, which makes it a good general template for problems involving level relationships. It directly uses Breadth-First Search on a Binary Tree.

Approach 2: Recursive Method (Time: O(n), Space: O(h))

The recursive strategy takes advantage of the perfect binary tree structure. For each node, connect node.left.next = node.right. Then connect across subtrees using node.right.next = node.next.left if the parent has a neighbor. Recursively apply the same logic to left and right children. The key insight: once a node's next pointer is set, it can be used to bridge connections across subtrees. This approach uses Depth-First Search and avoids an explicit queue.

Recommended for interviews: The BFS approach is the safest starting point because it works for both perfect and general binary trees. However, interviewers often expect the recursive solution for this specific problem since it leverages the perfect-tree constraint and reduces auxiliary space. Demonstrating BFS first shows correctness and clarity; transitioning to the recursive pointer-linking approach shows deeper understanding.