Populating Next Right Pointers in Each Node - Solution & Explanation

Problem Overview: Each node in a perfect binary tree has a next pointer that should point to its immediate neighbor on the right at the same level. If there is no neighbor, the pointer must be null. The task is to populate these pointers efficiently while traversing the tree.

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

This approach performs a level-by-level traversal using a queue, a standard pattern in breadth-first search. Push the root node into the queue, then process nodes one level at a time. For each level, iterate through the queue size and connect the current node’s next pointer to the next node in the queue unless it is the last node in that level. The queue naturally groups nodes by depth, making pointer connections straightforward. This method works for any binary tree structure and is easy to reason about during interviews.

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

This method leverages the properties of a perfect binary tree. Each node’s left child should point to its right child, and each node’s right child should point to the left child of the node’s next neighbor. Use depth-first search recursion to propagate these relationships down the tree. At every node, connect node.left.next = node.right. If the current node has a next, connect node.right.next = node.next.left. Recursively process the left and right subtrees. The recursion depth equals the tree height, which is log n for a perfect tree.

Recommended for interviews: Start with the BFS level order approach because it clearly demonstrates understanding of tree traversal and pointer linking. Interviewers often expect candidates to recognize that the tree is perfect and optimize further using the recursive constant-extra-space method. Showing both approaches demonstrates strong grasp of tree traversal patterns and pointer manipulation.