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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
For splaying, three trees are maintained, the central, left & right sub trees. At first, the central subtree is the complete tree and left and right subtrees are empty. The target
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Since the stack is list of elements, the queue is also a list of elements. The stack & the queue differ just in the position where the elements may be added or deleted. Similar to
Hubs - In reality a multiport repeater - Connects stations in a physical star topology - As well may create multiple levels of hierarchy to remove length limitation of 10
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Draw trace table and determine output from the subsequent flowchart using below data: X = 5, -3, 0, -3, 7, 0, 6, -11, -7, 12
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