Introduction

In this blog post, we will explore how to solve the LeetCode problem "513" 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 leftmost value in the last row of the tree. Example 1: Input: root = [2,1,3] Output: 1 Example 2: Input: root = [1,2,3,4,null,5,6,null,null,7] Output: 7 Constraints: The number of nodes in the tree is in the range [1, 104]. -231 <= Node.val <= 231 - 1

Explanation

Here's the breakdown of the solution:

• Level-Order Traversal: Use a level-order (breadth-first) traversal to visit nodes row by row.
• Track Last Row: Keep track of the leftmost node's value as we iterate through each level. Since we are traversing level by level, when the traversal is finished, we would've already reached the last level.

• Update Leftmost: For each new level, update the leftmost_value with the value of the first node encountered at that level.

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

• Space Complexity: O(W), where W is the maximum width of the tree (the maximum number of nodes at any level). In the worst case (a complete binary tree), W can be approximately N/2, so the space complexity can be considered O(N).

Code

    from collections import deque

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

def findBottomLeftValue(root):
    """
    Finds the leftmost value in the last row of a binary tree.

    Args:
        root: The root node of the binary tree.

    Returns:
        The leftmost value in the last row of the tree.
    """
    if not root:
        return None

    queue = deque([root])
    leftmost_value = 0

    while queue:
        level_size = len(queue)
        leftmost_value = queue[0].val  # Update leftmost for the current level

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

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

    return leftmost_value