Populating Next Right Pointers in Each Node II - Video Solutions

Problem Overview: You are given a binary tree where each node contains an extra next pointer. The task is to connect every node to its immediate neighbor on the same level. If there is no neighbor, the next pointer should be set to null. Unlike the simpler version of the problem, the tree is not guaranteed to be perfect, so children may appear in any structure.

Approach 1: Level Order Traversal Using a Queue (O(n) time, O(n) space)

This approach performs a standard Breadth-First Search level traversal. Push the root into a queue, then process nodes level by level. For each level, iterate through the nodes currently in the queue and connect the previous node's next pointer to the current node. Children of the current node are pushed into the queue so the next level can be processed afterward. Because every node is visited exactly once, the time complexity is O(n). The queue may hold up to the maximum width of the tree, giving O(n) worst-case space complexity.

This method is straightforward and mirrors how levels are naturally processed in a binary tree. It is often the first solution developers implement because the logic is simple: track the previous node in the level and link it to the current one.

Approach 2: Using Constant Space (Two Pointers) (O(n) time, O(1) extra space)

The optimized solution eliminates the queue by using the already established next pointers to traverse levels. Maintain two pointers: one for scanning the current level and another for building the next level. A temporary dummy node acts as the head of the next level while a tail pointer attaches children as they are discovered.

Start with the root as the current level. Iterate through nodes using their next pointers. Whenever a node has a left or right child, attach it to the tail of the next-level chain and advance the tail pointer. Once the current level is finished, move to dummy.next, which represents the start of the next level, and repeat the process.

This technique effectively performs a level traversal without extra storage. Each node is processed once, so the time complexity remains O(n). Since only a few pointers are used regardless of tree size, the auxiliary space complexity is O(1). The idea resembles pointer-based traversal patterns often seen in DFS style tree problems, but applied level-by-level.

Recommended for interviews: Start with the queue-based BFS explanation because it clearly demonstrates how level connections work. Then transition to the constant-space two-pointer approach. Interviewers typically expect the O(1) extra space solution since it shows strong understanding of pointer manipulation and tree traversal patterns.