Question

I am very confused by a number of articles at different sites regarding constructing a Binary Search Tree from any one traversal (pre,post or in-order), or a combination of any two of them. For example, at this page, it says that given the pre,post or level order traversal, along with the in-order traversal, one can construct the BST. But here and there, they show us to construct a BST from pre-order alone. Also, here they show us how to construct the BST from given pre and post-order traversals. In some other site, I found a solution for constructing a BST from the post-order traversal only.

Now I know that given the inorder and pre-order traversals, it is possible to uniquely form a BST. As regards the first link I provided, although they say that we can’t construct the BST from pre-order and post-order, can’t I just sort the post-order array to get its inorder traversal, and then use that and the pre-order array to form the BST? Will that be same as the solution in the 4th link, or different? And given pre-order only, I can sort that to get the in-order, then use that and the pre-order to get the BST. Again, does that have to be different from the solution at links 2 and 3?

Specifically, what is sufficient to uniquely generate the BST? If uniquement is not required, then I can simply sort it to get the in-order traversal, and build one of the N possible BSTs from it recursively.

Answer 1

To construct a BST you need only one (not in-order) traversal.

In general, to build a binary tree you are going to need two traversals, in order and pre-order for example. However, for the special case of BST – the in-order traversal is always the sorted array containing the elements, so you can always reconstruct it and use an algorithm to reconstruct a generic tree from pre-order and in-order traversals.

So, the information that the tree is a BST, along with the elements in it (even unordered) are equivalent to an in-order traversal.

Bonus: why is one traversal not enough for a general tree, (without the information it is a BST)?

Answer: Let’s assume we have n distinct elements. There are n! possible lists to these n elements, however – the possible number of trees is much larger (2 * n! possible trees for the n elements are all decayed trees, such that node.right = null in every node, thus the tree is actually a list to the right. There are n! such trees, and another n! trees where always node.left = null ) Thus, from pigeon hole principle – there is at least one list that generates 2 trees, thus we cannot reconstruct the tree from a single traversal.
(QED)