Solving Leetcode Interviews in Seconds with AI: Deepest Leaves Sum
Introduction
In this blog post, we will explore how to solve the LeetCode problem "1302" 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 sum of values of its deepest leaves. Example 1: Input: root = [1,2,3,4,5,null,6,7,null,null,null,null,8] Output: 15 Example 2: Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5] Output: 19 Constraints: The number of nodes in the tree is in the range [1, 104]. 1 <= Node.val <= 100
Explanation
Here's a solution to find the sum of deepest leaves in a binary tree, incorporating efficiency and optimization:
- Depth-First Search (DFS): Traverse the tree using DFS to determine the maximum depth.
- Conditional Summation: During the DFS, maintain the sum of node values only for nodes at the maximum depth found. Reset the sum if a deeper level is encountered.
Avoid Redundant Calculations: The single DFS ensures each node is visited only once, preventing redundant depth calculations or summations.
Runtime Complexity: O(N), Storage Complexity: O(H), where N is the number of nodes in the tree and H is the height of the tree (due to the call stack in DFS). In the worst-case (skewed tree), H can be N.
Code
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def deepestLeavesSum(root: TreeNode) -> int:
max_depth = 0
deepest_sum = 0
def dfs(node, depth):
nonlocal max_depth, deepest_sum
if not node:
return
if depth > max_depth:
max_depth = depth
deepest_sum = node.val
elif depth == max_depth:
deepest_sum += node.val
dfs(node.left, depth + 1)
dfs(node.right, depth + 1)
dfs(root, 0)
return deepest_sum