Introduction

In this blog post, we will explore how to solve the LeetCode problem "1104" 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

In an infinite binary tree where every node has two children, the nodes are labelled in row order. In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left. Given the label of a node in this tree, return the labels in the path from the root of the tree to the node with that label. Example 1: Input: label = 14 Output: [1,3,4,14] Example 2: Input: label = 26 Output: [1,2,6,10,26] Constraints: 1 <= label <= 10^6

Explanation

Here's the breakdown of the solution:

• Find the Row: Determine the row number in which the target label exists. The row number can be found by using the logarithm base 2.
• Path Reconstruction: Reconstruct the path from the target label back to the root (1) by iteratively finding the parent node's label. Since even rows are in reverse order, we need to adjust the parent label calculation based on the row number.

• Reverse for Root-to-Node Order: Since we computed the path from the target to the root, reverse the resulting path to get the desired root-to-node order.

• Runtime Complexity: O(log N) where N is the input label, and Storage Complexity: O(log N) for storing the path.

Code

    import math

def pathInZigZagTree(label: int) -> list[int]:
    """
    Given the label of a node in this tree, return the labels in the path from the root
    of the tree to the node with that label.
    """

    path = [label]
    row = int(math.log2(label))
    while label > 1:
        parent = label // 2
        if (row - 1) % 2 != 0:
            start = 2**(row - 1)
            end = 2**row - 1
            parent = start + end - parent
        path.append(parent)
        label = parent
        row -= 1

    return path[::-1]