递归和迭代DFS算法时间复杂度和空间复杂度

Question

So I am having some problems understanding why the time complexity of a recursive DFS and an iterative DFS is the same, perhaps someone can guide me through an easy explanation?

Thanks in advance.

Answer 1

A recursive implementation requires, in the worst case, a number of stack frames (invocations of subroutines that have not finished running yet) proportional to the number of vertices in the graph. This worst-case bound is reached on, eg, a path graph if we start at one end.

An iterative implementation requires, in the worst case, a number of stack entries proportional to the number of vertices in the graph. The same inputs reach this worst-case time as for the recursive implementation.