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

Populating Next Right Pointers in Each Node II (LeetCode #117) 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: Using next pointers can help reduce space complexity.

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

The brute force approach for Populating Next Right Pointers in Each Node II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the tree level by level and connecting each node to its immediate right neighbor. This is done by using a nested loop to find the next right node for each node in the current level. The optimal Optimal Solution approach improves this to O(n).

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

Always clarify constraints regarding space usage. Discuss your thought process and approach before coding. Test your solution with edge cases, like empty trees.

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

Forgetting to handle NULL cases for children nodes. Not resetting pointers correctly between levels.

#117

Populating Next Right Pointers in Each Node II

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)

The optimal solution uses a constant space approach by leveraging the next pointers already established. We traverse the tree level by level, using the next pointers to connect nodes without needing additional data structures.

⚙️

Algorithm

3 steps

1Step 1: Start from the root and initialize a pointer to track the current node at the current level.
2Step 2: For each node, connect its left and right children to the next node in the same level using the next pointers.
3Step 3: Move to the next level using the established next pointers.

solution.py30 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:
14        curr = leftmost
15        leftmost = None
16        prev = None
17        while curr:
18            if curr.left:
19                if prev:
20                    prev.next = curr.left
21                else:
22                    leftmost = curr.left
23                prev = curr.left
24            if curr.right:
25                if prev:
26                    prev.next = curr.right
27                else:
28                    leftmost = curr.right
29                prev = curr.right
30            curr = curr.next

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(1) because we are using only a constant amount of extra space, leveraging the existing next pointers.

1Using next pointers can help reduce space complexity.
2Level-order traversal is key to connecting nodes in the same level.

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