Given the head of a singly linked list and an integer target, return true if target appears anywhere in the list, and false otherwise. Unlike an array, a linked list gives no random access — the only way to reach any node is by walking next pointers from head, so a linear scan is unavoidable, sorted or not. Both an iterativeIterativeSolved with an explicit loop, keeping state (like a curr pointer) in local variables instead of the call stack. Runs in O(1) extra space here. pass with a curr pointer and a recursiveRecursiveSolved by having the function call itself on a smaller version of the same problem — here, "search the rest of the list" is the same problem as the original, just one node shorter. version that checks head and calls itself on head.next do exactly this walk; the recursive version just spends O(n) call-stack space doing it.