Is Populating Next Right Pointers in Each Node II easy or hard?

The problem is rated Medium on LeetCode. The BFS solution is straightforward, but the constant-space solution requires careful pointer handling across levels, which increases the difficulty slightly.

Populating Next Right Pointers in Each Node II Python/Java solution

Python and Java implementations usually follow either BFS with a queue or the constant-space pointer technique. Both versions iterate through levels, connect adjacent nodes, and ensure the last node in each level points to null.

How to solve Populating Next Right Pointers in Each Node II in O(n)?

Traverse the tree level by level. During traversal, connect nodes within the same level either by storing nodes in a queue or by using existing next pointers to iterate horizontally. Each node is processed one time, giving O(n) time complexity.

What is the best approach for Populating Next Right Pointers in Each Node II?

The most efficient approach uses a constant-space two-pointer technique. It traverses each level using existing next pointers while building the next level with a dummy head and tail pointer. This processes all nodes in O(n) time and O(1) extra space, making it the preferred interview solution.

Is Populating Next Right Pointers in Each Node II asked at Google/Amazon/Meta?

Tree pointer manipulation and level traversal problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variants of this problem are commonly used to evaluate understanding of BFS traversal and pointer-based tree techniques.

What data structure is used in Populating Next Right Pointers in Each Node II?

The primary data structure is a binary tree. Solutions typically use a queue for Breadth-First Search or pointer manipulation to traverse levels without additional storage. The optimized approach relies on linking nodes using the existing next pointers.

What is the time complexity of Populating Next Right Pointers in Each Node II?

Both common approaches run in O(n) time because every node in the binary tree is visited exactly once. The queue-based BFS uses O(n) space in the worst case, while the optimized two-pointer solution reduces the extra space to O(1).

Description Solution Video 8:16

117. Populating Next Right Pointers in Each Node II

Medium57.9% acceptance7 companies

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 6000].
-100 <= Node.val <= 100

Follow-up:

You may only use constant extra space.
The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.