Why is the time complexity of DFS to detect a cycle in an undirected graph O(|V|) and not O(|V| + |E|)?

Question

Can anyone explain to me in detail why and how the upper bound of DFS to detect a cycle in an undirected graph be O(|V|) ?

Answer 1

A graph without cycles has at most |V| - 1 edges (it's a forest). Therefore if the DFS discovers |V| edges or more then it already found a cycle and terminates. The runtime is accordingly bounded by O(|V|).