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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
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Q. Explain that how do we implement two stacks in one array A[1..n] in such a way that neither the stack overflows unless the total number of elements in both stacks together is n.
How to measure the algorithm's efficiency? It is logical to examine the algorithm's efficiency as a function of some parameter n showing the algorithm's input size. Instance
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