What is the number of distinct BFS, DFS in a complete graph?

My approach is like this-->

For DFS take a node, we have (n-1) possible tree edges, next we have all possible (n-2) tree edges,.......finally 1 possible tree edge. and we can choos any of n nodes as rooot node. hence no of distinct DFS trees= n*(n-1)(n-2)....1= n!

For BFS tree if we take a node as root, we have to explore all its neighbors , so the number distinct of BFS trees =no of nodes we can choose as root ie. n .

Is this approach correct?

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  • $\begingroup$ Please help... :( $\endgroup$ – Turing101 Aug 11 '19 at 4:40

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