I came to know through some examples that DFS and Backtracking aren't exactly the same ( A misconception I had since a long time). So now my question is, since Backtracking visits nodes backwards step by step while DFS on a graph may directly jump some nodes backwards, is DFS a faster algorithm? If so, only in terms of constants or complexity wise?
Edit: Maybe the original writeup was a bit confusing. What I mean specifically is this.
A normal implemetation of DFS should traverse the following tree in this manner A->B->C->D. This implies that the implementation for what we normally refer to as DFS visits each node exactly once, giving it a time complexity O(V) where V is the number of vertices.
This I feel is a more bare bones approach where one step of backtracking actually goes to a parent node and explores its other children, instead of skipping over some parent nodes which had only one child (node B here) to begin with. This results in a traversal sequence of A->B->C->B->A->D. This particular example shows that this is always going to require more nodes to be visited than the standard DFS and hence the question.