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CHAPTER 14 TREE-STRUCTURED DATA
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operator inorder ( root ) ; emit inorder ( left child ( root ) ) ; emit root ; emit inorder ( right child ( root ) ) ; end inorder ; Note: I ll explain the reason for the name inorder in the next section. So how do we build the tree in other words, how do we do the actual sort Well, that s a recursive process too. Pseudocode: tree := empty ; i := 0 ; do i := i + 1 until no more input values ; if i = 1 then insert ith value into tree ; /* root */ else call add_to_tree ( root ) ; end if ; end do ; operator add_to_tree ( root ) ; if ith value root value then if root has no left child then insert ith value as left child ( root ) ; else call add_to_tree ( left child ( root ) ) ; end if ; else if root has no right child then insert ith value as right child ( root ) ; else call add_to_tree ( right child ( root ) ) ; end if ; end if ; end add_to_tree ;
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Treesort clearly involves three data structures: the input list, the output list, and the tree. How can we represent these structures relationally Well, the input and output lists are easy: INLIST { P INTEGER, V INTEGER } KEY { P } OUTLIST { P INTEGER, V INTEGER } KEY { P } Explanation: If there are n values (integers, by our earlier assumption) to be sorted, INLIST and OUTLIST each contain or will contain n tuples, and attribute P ( position ) takes or will take exactly the values 1, 2, ..., n. For INLIST, P = i corresponds to the value appearing in the ith position in the input. For OUTLIST, P = i corresponds to the value appearing in the ith position in the sorted result.
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CHAPTER 14 TREE-STRUCTURED DATA
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What about the tree One simple design is: TREE { ID INTEGER, V INTEGER, LEFT INTEGER, RIGHT INTEGER } KEY { ID } Explanation: After it s built, the tree will contain n nodes i.e., there will be n tuples in TREE and attribute ID ( node identifier ) will take exactly the values 1, 2, ..., n. For a given node, LEFT and RIGHT refer to the left and right child, respectively; I adopt the convention that a LEFT or RIGHT value of zero means no corresponding child exists. Note: As I ve explained elsewhere (see, e.g., my book An Introduction to Database Systems, 8th edition, Addison-Wesley, 2004), INLIST, OUTLIST, and TREE are strictly speaking not relations as such but, rather, relation variables (relvars for short). From this point forward I ll be careful always to use the term relvar when it really is a relvar I m talking about, and reserve the term relation to mean a relation value (in particular, the value of some relvar).
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Essentially I want to propose just one new relational operator here, which I ll call (unsurprisingly) TREESORT. TREESORT takes a single relation as input and produces another as output. The input relation has two attributes, which I ll assume for the moment are called ID and V and have semantics as explained in the previous subsection, and the output relation has the same heading as the input one. Thus, if INLIST is as in the previous subsection, the expression TREESORT ( INLIST ) yields a relation looking like, and having the same semantics as, OUTLIST in the previous subsection. Points arising: In practice we would need a way to specify which attribute is which, as it were, in the input relation, so that TREESORT knows which is the identifier attribute and which the value one. I omit such details here for simplicity. In practice we wouldn t want TREESORT to be limited to sorting lists of integers but to work for values of any ordinal type (where an ordinal type is simply a type for which the operator < is defined). I limit my attention to integers here for simplicity. It should be clear that the user of TREESORT needs to know the structure of the input and output relations but not the structure of the tree relation, which is purely internal to the TREESORT implementation. Of course, we could make the structure of that tree relation externally visible (and provide user-visible BUILD and INORDER operators as well) if we had some good reason to do so. One final point: Remarks similar to those of the present subsection apply to the discussions of the next section ( Traversing the Nodes of a Tree ) also, mutatis mutandis, and I won t bother to spell the details out again in that section.
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