Introduction

In this blog post, we will explore how to solve the LeetCode problem "107" 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

Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root). Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[15,7],[9,20],[3]] Example 2: Input: root = [1] Output: [[1]] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 2000]. -1000 <= Node.val <= 1000

Explanation

Here's the breakdown of the problem and the solution:

• High-Level Approach:

  • Use Breadth-First Search (BFS) to traverse the tree level by level.
  • Store the nodes of each level in a temporary list.
  • Append each level's list to the beginning of the final result list, effectively reversing the level order.

• Complexity Analysis:

  • Runtime Complexity: O(N), where N is the number of nodes in the tree.
  • Storage Complexity: O(W), where W is the maximum width of the tree (maximum number of nodes at any level). In the worst case (a complete binary tree), W can be close to N, so the space complexity can be considered O(N).

Code

    from collections import deque
from typing import List, Optional

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []

        result = []
        queue = deque([root])

        while queue:
            level_size = len(queue)
            current_level = []

            for _ in range(level_size):
                node = queue.popleft()
                current_level.append(node.val)

                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            result.insert(0, current_level)  # Prepend to reverse the order

        return result