Leetcode considers this question to be easy, but it was a bit more complicated than that at first for me. Let’s try solving it three different ways:
Whenever I can, I always try to do BFS. It seems much more natural.
BFS - 1st solution func isSymmetricBFS(_ root: TreeNode?) -> Bool { var queue: [TreeNode?] = [] queue.append(root) while !queue.isEmpty { let temp = queue queue.removeAll() // 🔑 to make sure you don't expand on `nil` nodes — otherwise it would be endless for node in temp where node !...
I do leetcoding every once a while, but keep forgetting some tree basics. This post here it to help with that.
Node vs Side 💡 This was a very subtle yet “Aha” moment for me. When you’re traversing down a tree using dfs, while you can do things like:
let leftNode = process(node.left) let rightNode = process(node.right) it might be better to see at as:
let leftSide = process(node.left) let rightSide = process(node....
My team was dealing with a large flow, where user can transition from multiple states or sometimes skip certain states. We didn’t have a centralized controller, every screen just had logic on where it should go next.
This made it difficult for us to see all our logic at once. We asked around and was told state machines are a good fit for our situation.
State machines are void of any UX....