Introduction

In this blog post, we will explore how to solve the LeetCode problem "116" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: 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 = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect 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, 212 - 1]. -1000 <= Node.val <= 1000 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.

Explanation

• Level-Order Traversal with Constant Space: The core idea is to perform a level-order traversal without using a queue (which would require O(N) space). We leverage the existing next pointers from the previous level to traverse the current level.
  • Iterating through Levels: We start at the root and, for each level, use the leftmost node of that level to iterate through all nodes on that level using the next pointers.
  • Connecting Children: While iterating through each level, we connect the left and right children of each node to their appropriate next nodes on the next level.

• Time Complexity: O(N), where N is the number of nodes in the tree. Space Complexity: O(1) (constant space).

Code

    class Node:
    def __init__(self, val=0, left=None, right=None, next=None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next

def connect(root: 'Node') -> 'Node':
    if not root:
        return None

    leftmost = root  # Leftmost node of the current level

    while leftmost:
        curr = leftmost
        leftmost = None  # Reset for the next level

        while curr:
            if curr.left:
                if not leftmost:
                    leftmost = curr.left
                curr.left.next = curr.right

            if curr.right:
                if curr.next:
                    curr.right.next = curr.next.left
            curr = curr.next

    return root