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In this respect depth-first search (DFS) is the exact reverse process: whenever it sends a new node, it immediately continues to extend from it. It sends back to previously explored nodes only if it lay out of options. Although DFS goes to unbalanced and strange-looking exploration trees related to the orderly layers created by BFS, the combination of eager exploration with the perfect memory of a computer creates DFS very useful. It sends an algorithm template for DFS. We send special algorithms from it by specifying the subroutines traverseTreeEdge, root, init, backtrack, and traverseNonTreeEdge.
DFS creates a node when it First discovers it; started all nodes are unmarked. The main loop of DFS seems for unmarked nodes s and calls DFS(s; s) to lead a tree rooted at s. The genuine call DFS(u; v) extends all edges (v;w) out of v. The argument (u; v) display that v was reached via the edge (u; v) into v. For root nodes s, we need the .dummy. argument (s; s). We display DFS(¤; v) if the special nature of the incoming node is irrelevant for the discussion at hand. Assume now that we explore edge (v;w) within the fact DFS(¤; v). If w has been seen after, w is a node of the DFS-tree. So (v;w) is not a tree node and hence we create traverseNonTreeEdge(v;w) and prepare no recursive call of DFS. If w has not been given before, (v;w) converts a tree edge. We therefore call traverseTreeEdge(v;w), mark w and create the recursive call DFS(v;w). When we return from this call we include the next edge out of v. Once all edges out of v are included, we call backtrack on the incoming edge (u; v) to operate any summarizing or clean-up operations return and required.
Which sorting algorithm is best if the list is already sorted? Why? Insertion sort as there is no movement of data if the list is already sorted and complexity is of the order
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The time needed to delete a node x from a doubly linked list having n nodes is O (1)
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Write a program that uses the radix sort to sort 1000 random digits. Print the data before and after the sort. Each sort bucket should be a linked list. At the end of the sort, the
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extra key inserted at end of array is called
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