🎀LeetCode Problem 2467🎀 Most Profitable Path in a Tree

🎀LeetCode Problem 2467🎀 Most Profitable Path in a Tree

🎀 : Alright besties, let's break down this tree problem! First, we gotta understand what's being asked. We're given a tree, Alice and Bob, and some amounts at each node. Alice wants to maximize her profit going to a leaf. Bob goes to 0. If they meet, they split the reward. Let's setup the function and adjacency list. We'll come back and fill in the core logic later, but for now, we'll just initialize the adjacency list using the given edges. Like, literally just convert the edges into an adjacency list. Capiche?
🎀 : Okay, now for the juicy part: the core logic! First, let's do a DFS to build the `parent` array, so we know how Bob gets back to node 0. Then, we calculate `bobReachTime` for each node. Think of it like this: We start Bob at his node and work backwards, tracking how many moves it takes him to get to each node. Finally, we use another DFS (`dfs_sum`) to find Alice's most profitable path. Alice gets the full amount if she reaches a node before Bob, half if they arrive together. This `dfs_sum` recursively explores each path to a leaf, maximizing Alice's income. Get it? It's like a race against time and Bob!
🎀 : Okay, queens and kings, we've got the core logic down! Now, let's refine this code to make it ✨sparkle✨. I've added comments to explain each part, so future you (or another dev) isn't totally lost. Error handling isn't super crucial here since the constraints guarantee a valid tree, but good comments and clean code are ALWAYS in style. Also, I cleared the adjacency list in `mostProfitablePath` to avoid potential issues if we were calling this function multiple times (although LeetCode doesn't do that here). Optimizations? We *could* potentially optimize the DFS with memoization, but honestly, it's already pretty efficient. The important thing is that it's readable, understandable, and works! Yas!

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