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We have discussed already about three tree traversal methods in the earlier section on general tree. The similar three different ways to do the traversal -inorder , preorder, and postorder are applicable to binary tree also.
Let us discuss the inorder binary tree traversal for given binary tree:
We begin from the root i.e. * we are assumed to visit its left sub-tree then visit the node itself & its right sub-tree. Here, root contain a left sub-tree rooted at +. Thus, we move to + and verify for its left sub-tree (we are supposed repeat this for each node). Again, + contain a left sub-tree rooted at 4. Thus, we need to check for 4's left sub-tree now, however 4 doesn't have any left sub-tree and therefore we will visit node 4 first (print in our case) and verify for its right sub-tree. As 4 doesn't contain any right sub-tree, we'll go back & visit node +; and verify for the right sub-tree of +. It contains a right sub-tree rooted at 5 and thus we move to 5. Well, 5 don't have any left or right sub-tree. Thus, we just visit 5 (print 5) and track back to +. As we already have visited + thus we track back to * . As we are yet to visit the node itself and thus we visit * before checking for the right sub-tree of *, which is 3. As 3 do not have any left or right sub-trees, we visit 3 . Thus, the inorder traversal results in 4 + 5 * 3
Breadth-first search starts at a given vertex h, which is at level 0. In the first stage, we go to all the vertices that are at the distance of one edge away. When we go there, we
In the array implementation of lists, elements are stored into continuous locations. In order to add an element into the list at the end, we can insert it without any problem. But,
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One can change a binary tree into its mirror image by traversing it in Postorder is the only proecess whcih can convert binary tree into its mirror image.
what is Paging.
How can we convert a graph into a tree ? Do we have any standardized algorithm for doing this?
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Ans: A procedure to reverse the singly linked list: reverse(struct node **st) { struct node *p, *q, *r; p = *st; q = NULL; while(p != NULL) { r =q;
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