How do you solve Populating Next Right Pointers in Each Node on LeetCode?

Populating Next Right Pointers in Each Node (LeetCode #116) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The tree is perfect, meaning all levels are fully filled, which allows for direct connections.

What is the brute force solution for Populating Next Right Pointers in Each Node?

The brute force approach for Populating Next Right Pointers in Each Node is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we traverse the tree level by level and for each node, we find its next right node by checking all nodes at the same level. This is simple but inefficient as it requires multiple passes over the tree. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Populating Next Right Pointers in Each Node?

Clarify the structure of the tree before starting to code. Discuss your thought process and approach with the interviewer. Be mindful of space complexity constraints when discussing your solution.

What are common mistakes when solving Populating Next Right Pointers in Each Node?

Not recognizing the tree is perfect, leading to unnecessary complexity. Confusing the next pointers with child pointers.

#116

Populating Next Right Pointers in Each Node

Medium

Linked ListTreeDepth-First SearchBreadth-First SearchBinary TreeTwo PointersLevel Order Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

This approach leverages the perfect structure of the binary tree to connect nodes using existing pointers without additional space. We use a pointer to traverse the current level and connect nodes directly.

⚙️

Algorithm

3 steps

1Step 1: Start from the root node and set a pointer to the leftmost node of the current level.
2Step 2: While there are nodes at the current level, connect the left and right children of each node.
3Step 3: Move the pointer to the next node at the same level until all nodes are processed.

solution.py21 lines

1# Full working Python code
2class Node:
3    def __init__(self, val=0, left=None, right=None, next=None):
4        self.val = val
5        self.left = left
6        self.right = right
7        self.next = next
8
9def connect(root):
10    if not root:
11        return
12    leftmost = root
13    while leftmost.left:
14        current = leftmost
15        while current:
16            current.left.next = current.right
17            if current.next:
18                current.right.next = current.next.left
19            current = current.next
20        leftmost = leftmost.left
21

ℹ

Complexity note: The time complexity is linear because we visit each node exactly once. The space complexity is constant since we use only a few pointers and do not utilize additional data structures.

1The tree is perfect, meaning all levels are fully filled, which allows for direct connections.
2Using existing pointers to connect nodes saves space and improves efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.