university of 
 
 illinois library 
 
 At urbana-champaign 
 
Digitized by the Internet Archive 
 in 2013 
 
 http://archive.org/details/generatingbinary888zaks 
 
7ir, 777 
 
 7^r/ 
 
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 UIUCDCS-R-77-838 
 
 UILU-ENG 77 17^8 
 
 GENERATING BINARY TREES LEXICOGRAPHICALLY 
 by 
 
 S. Zaks 
 
 August 1977 
 
 DEPARTMENT OF COMPUTER SCIENCE 
 UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 
 
 URBANA, ILLINOIS 
 
 970 
 
GENERATING BINARY TREES LEXICOGRAPHICALLY* 
 
 by 
 S. Zaks 
 
 Department of Computer Science 
 University of Illinois at Urbana-Champaign 
 Urbana, Illinois 61801 
 
 August 1977 
 
 * 
 
 This work was supported in part by the National Science Foundation 
 under Grant Number MCS-73-03408. 
 

 TABLE OF CONTENTS 
 
 1. Introduction 
 
 2. Definitions and Notations 
 
 3. Existing Algorithms 
 
 4. Trees and Integer Sequences; The Generating Algorithm 
 
 5. The Ranking Function 
 
 6. The Unranking Procedure 
 
 7. k-ary Trees 
 
 8. Summary 
 
 Appendix: An example to the structures and sequences mentioned in the 
 paper. 
 
I. INTRODUCTION 
 
 We show a 1-1 correspondences among all the regular binary trees 
 
 On 
 
 with n internal nodes, all the 0,1-sequences x = {x.}, of n l's and n O's 
 in which the total number of l's in each prefix {x.}-, [i <_ n) is at least 
 as the total number of O's, all the integer sequences y = {y.K in which 
 y,<y2< ... ^y^n and y.^2i for i = l,2,...,n, and all the integer sequences 
 z = {z . }, in which < z-, < z ? < ... < z and z. <_ 2i - 1 for i = 1 , 2, . . . , n, 
 
 The term "lexicographic ordering of trees" is discussed, and the 
 relation between it and the lexicographic ordering of the corresponding 
 sequences is shown. 
 
 For these sequences we (1) develop an algorithm which generates 
 them lexicographically, (2) show how to find the position of a given sequence 
 in the lexicographic ordering, and (3) show how to find a sequence given its 
 position. 
 
 We prove (Theorem 4.8) that our order of generating these trees is 
 equivalent to the order in which they are generated by two existing algorithms. 
 The generating procedure is then generalized for the k-ary case. 
 
 Section 2 introduces basic definitions and notations, including two 
 different definitions for "lexicographic ordering of trees." In Section 3 
 we discuss existing related algorithms. Section 4 establishes the 1-1 
 correspondences mentioned above, and algorithm which generates binary trees is 
 presented. Sections 5 and 6 deal with the ranking function and the unrank- 
 ing procedure, respectively. In Section 7 we generalize the generating 
 algorithm for k-ary trees. Additional comments and a summary of our dis- 
 cussions are presented in Section 8. A list of all the regular binary trees 
 with 4 internal nodes and the associated corresponding structures and sequence 
 are shown in the Appendix. 
 
II. DEFINITIONS AND NOTATIONS 
 
 2.1 We state here defintions and notations that we shall use throughout this 
 paper. We follow mainly the terminologies in [KN 1] and [LI 2]. 
 
 2.2 While saying tree we mean an ordered tree . For k >_ 2 we define a k-ary 
 tree T as follows: either T is empty or it has a distinguished node r called 
 its root that is connected to T, , T ? , ... , T. , each of which is a k-ary tree. 
 For k = 2 we write T. and T R instead of T, and T ? , respectively, which will be 
 referred to as left subtree and right subtree . The root of each non-empty 
 
 T. for 1 <_ i <_ k is called a son of the root r, while r is its father . A 
 vertex in a tree can be an internal node or an external node (also called a 
 leaf ). A regular k-ary tree is a non-empty k-ary tree with each internal node 
 having k sons. We denote by |T| the number of vertices of a tree T. We denote 
 T(k,n ) = {all the regular k-ary trees with n internal nodes}. t(k,n) will be 
 the number of elements in T(k,n). We define B = T(2,n) and b = t(2,n). 
 
 2.3 There is a well -known 1-1 correspondence between B and all the binary 
 
 trees with n vertices: From a tree T e B remove all the leaves and the edges 
 
 n 
 
 incident with them. We get a binary tree with n vertices, and this trans- 
 formation establishes the 1-1 correspondence. (See [KN 1: p. 559] for this 
 correspondence, and [TR 1: p. 3] for an extension for k-ary trees). 
 
 2.4 We explain first what we mean by a "lexicographic order" of trees. The 
 following definition is used in [KN 1: 2.3.1, ex. 25], [KNO: p. 113] and the 
 first few sections of [TR 1]: 
 
 Definition 2.5 : Given two k-ary trees T and T', we say that T < T' if 
 
 (1) |T| < |T'|. or 
 
 (2) |T| = |T' | > and for some 1 <_ i <_ k we have 
 (a) T. = T. for j = 1 , 2, ... , i-1, and 
 
 (b) t, < t: 
 
(See [TR 1: p. 16]). This defines a linear order in T(k,n). See Appendix 
 for an example. 
 
 2.6 Unfortunately, no algorithm is known yet which generates those trees 1n 
 order using this definition. In [KNO] and [TR 1] we have the ranking func- 
 tion and unranking procedure .which may be used in an indirect way to generate 
 the trees lexicographically (given a tree, apply the ranking function to find 
 its position, then apply the unranking procedure to build the tree that occupies 
 the next position). 
 
 2.7 We define now another "lexicographic order" of trees. As will be shown 
 later, all the existing algorithms which (directly ) generate trees in 
 "lexicographic order" are using this definition. 
 
 Definition 2.8 : Given two k-ary trees T and T', we say that T < V if 
 
 (1 ) T is empty, or 
 
 (2) T is not empty, and for some i, 1 <_ i <_ k, we have 
 
 (a) T, = T. for j = 1 , 2, . .. , i-1, and 
 
 (b) T. <t'.. 
 
 This defines a linear order in T(k,n). See Appendix for an example. 
 
 2.9 As explained in 2.6, we use definition 2.8 throughout this work. The two 
 
 definitions 2.5 and 2.8 are not equivalent. For example, T < T' but T' < T for 
 
 (2.8) (2.5) 
 the trees T and T 1 in Figure 1. 
 
 Ordering of Trees 
 Figure 1 
 
2.10 We know that 
 
 In particular, 
 
 b n = nTr( 2 n n )' 
 (See [KN 1: p. 584]). 
 
 2.11 Our discussion contains three parts: 
 
 (1) Generating step : we generate B lexicographically (accord- 
 ing to the ordering as discussed above). 
 
 (2) Ranking step : we compute the function Index (T) , which assigns 
 to a given tree its appropriate position in this ordering. 
 
 (3) Unranking step : given a position x in this linear ordering, 
 we construct the tree T s.t. Index (T) = x. 
 
 Step (1) is then generalized for an arbitrary k. 
 
III. EXISTING ALGORITHMS 
 
 The following results discuss some or all the steps of 2.11: 
 
 3.1 In [RH] a binary tree is represented by the sequenceof the level numbers of 
 its leaves from left to right. Those sequences are then generated lexico- 
 graphically (we show here - Theorem 4.8 - that the corresponding trees are 
 generated thus in order according to definition 2.8.), and the ranking and 
 unranking functions are discussed. This way of labeling can be extended to 
 k-ary trees (see Theorem 7.8). 
 
 3.2 In [KN 1: 2.2.1, ex. 2,4,5; 2.3.1 ex. 6], [TR 1, 2] and [KNO] the 
 numbers 1, 2, ... , 2n+l are used to label the vertices of a tree Te B in 
 some order (say, preorder: Root-Left-Right) and then these labels are read in 
 another order (say, inorder: Left-Root-Right). We get a permutation of(l, 2, 
 ... , 2n+U. The preorder-inorder choice (used above in the brackets) is used 
 in [TR 1,2], in which all the three steps of generating, ranking and unranking 
 are discussed. Generalizations of the ranking and of the unranking steps for 
 k-ary trees are discussed in [TR 1,2]. All those results use definition 1.8 
 for "lexicographic order" (see theorems 3.8 and 6.8). The various possible 
 label ings are discussed in [KNO], which deals with ranking and unranking for 
 binary trees, using definition 1.5. This definition is used - for ranking 
 and unranking in the k-ary case - in [TR 1]. 
 
 3.3 For the two labeling sequenses mentioned above, see example in the Appendix, 
 
IV. TREES AND INTEGER SEQUENCES; THE GENERATING ALGORITHM 
 
 4.1 The number b of regular binary trees with n internal nodes is the 
 
 n-th Catalan number, b = —rr I „ )• Given such a tree T, we label each inter- 
 
 n n+1 \n J 
 
 nal node with 1 and each leaf with 0. We then read these labels in preorder 
 
 (Root-Left-Right). We get a sequence of n l's and n+1 0's. As the last 
 
 visited node is always a leaf, we omit the corresponding 0. Denote the 
 
 resulting sequence by f(T) = {x.} = x. From x we build a sequence g(T)= 
 
 1 1 
 
 iy.r = y s.t. 
 1 1 
 
 y. = position of the i-th in x 
 and a sequence h(T) = {z.}, = z s.t. 
 
 z. = position of the i-th 1 in x. 
 When no confusion occurs, we omit commas and brackets in writing a sequence 
 explicitly. See Figure 2 for an example. 
 
 x = 1111000101100100 
 
 y = 5, 6, 7, 9, 12, 13, 15, 16 
 
 z = 1, 2, 3, 4, 8, 10, 11, 14 
 
 Labeling a Binary Tree 
 Figure 2 
 A sequence x = {x.} -j will be called feasible if there is a tree T 
 s.t. x = f(T). The corresponding definition holds for the y and z sequences. 
 
 A 0,Vseuqnce is said to have the dominating property if in any 
 prefix the number of l's is at least as the number of 0's. 
 
p_- 
 
 Theorem 4.2: A 0,1-sequence x = { x .} -, is feasible iff 
 
 (1) It has n 1 ' s and n O's, and 
 
 (2) It has the dominating property. 
 
 Proof : If x is feasible then (1) holds clearly. For (2) we note that the 
 number of internal nodes we visit, as described in 4.1, is always at least 
 as the number of leaves (except in the last step, when we have visited n + 1 
 leaves but only n internal nodes; but this last leaf was excluded in x). 
 
 These two conditions are also sufficient, because the number of those 
 sequences equal the number of those trees (see 4.4), and two different trees 
 T and T' correspond to two different sequences x and x 1 by Theorem 4.8. n 
 
 4.3 It should be mentioned that associating the sequence f(T) with T e B can 
 be found in other references (see [GA]; and [KN 3: p. 63] for a related label- 
 ing). The correspondence between x = f(T) and z = h(T) is mentioned in [KN 3: 
 p. 63], using the notion of Young's tableau. Our contribution is in making 
 use of these sequences for the generating, ranking and unranking procedures 
 and in generalizing these sequences and the generating procedure for k-ary 
 trees. 
 
 4.4 The box-office problem is usually related to those o,l~sequences (see 
 [YY: problem 83], [GN: p. 36]), with the solution ~j ( 2 n n ). Other interpreta- 
 tions of these sequences, such as Bertrand's balloting problem, are known 
 (see [LI 1: p. 74], [GA]). 
 
 4.5 The following theorem establishes the relation between binary trees and 
 integer sequences: 
 
 Theorem 4.6: The following sets are in a 1-1 correspondence with one another: 
 
 (1) B 
 v ' n 
 
 The box-office problem: 2n people are waiting in a line at a box office, 
 n of them have 1 dollar bills, and the rest have 2 dollar bills. Tickets cost 
 1 dollar each. When the box-office opens, there is no money in the till. Each 
 customer buys one ticket. In how many ways can they stand such that none of them 
 will have to wait for change? 
 
8 
 
 2n 
 
 (2) All the 0,1-sequences x = (x-}, , with n l's and n O's, having 
 
 the dominating property. 
 
 n 
 
 (3) All the integer sequences y = {y,-}-i> s.t. y-i < y? < • • • < y < 
 
 2n and y. >_ 2i for i = 1 , 2, . . . , n. 
 
 n 
 
 (4) All the interger sequences z = {z,-}-i. s.t. < z, < z« < ... < 
 
 z and z . <_ 2i - 1 for i = 1 , 2, . > . , n. 
 
 Proof : (1) e (2) 
 
 See Theorem 4.2. 
 
 (2) e (3) 
 
 2n 
 With a 0,1-sequence x = {x.}, , we associate a corresponding sequence 
 n n 
 
 {y.}, as explained in 4.1. 
 
 Suppose x satisfies (2). If for some £, 1 <_ £ <_ n, y < 2£, then 
 among x-, , x 2 , ... , x ?0-i there are at least £ O's, which violates the dominat- 
 ing property of x. Hence y. >_ 2i for i = 1, 2, ... , n. 
 
 Conversely, assume x doesn't satisfy (2). It it doesn't have n O's 
 
 its corresponding y(x) doesn't satisfy (3). If it has n l's and n O's but 
 
 doesn't have the dominating property, then for some £, 1 <_ z <_ 2n, the num- 
 
 z 
 ber t of O's in a prefix {x-K is more than the number I - t of l's, or 
 
 i < 2t. Hence the position y. of the t-th staisfies y. <_ £ < 2t , hence y 
 
 doesn't satisfy (3). 
 
 It is clear that different sequences x and x' correspond to different 
 
 sequences y and y', respectively, and this completes the proof. 
 
 (3) = (4) 
 Left to the reader. □ 
 
 n 
 4.7 We define the lexicographic order for sequences as usual: Let u = {u^ }, 
 
 n 
 and v = {v.},. We say that u < v if there exist an index £, 1 <_ l <_ n, s.t. 
 
 u. = v. for j = 1, 2, ... , £-1 and u < v . 
 
 J J J6 X. 
 
 The following observation is essential for our work: 
 
 
Theorem 4.8 : Let T and T' be two regular binary trees with n internal nodes, 
 and let the corresponding sequences x, y, z and x', y', z' be as defined above. 
 
 The following are eauivalent: 
 
 (1) T < T' (definition 2.8) 
 
 (2) x < x' 
 
 (3) y < y' 
 
 (4) z > z' 
 
 (5) T < T' according to Ruskey-Hu's level sequence (see 3.1) 
 
 (6) T < T' according to Trojanowski 's permutation (see 3.2) 
 Proof: (2) ~ (3) ~ (4) 
 
 Immediately from the definition of y and z (4.1). 
 
 (1) ~ (2) 
 
 T < T' iff the i vertex visited is a leaf in T but is an internal 
 node in T', while the first i - 1 vertices are the same, for some i, 1 < i < 
 2n. This happens iff x. = x. for i = 1 , 2, . . . , £-1 and x = < 1 = x 
 
 X/ X/ 
 
 for some i as above, which happens iff x < x 1 . 
 
 (2) ~ (5) 
 
 x < x' iff for some i, !<£<2n, x i = x^ for i = 1 ,2, . . . , £-1 , and x £ = 0<1 = 
 x . Let us look at those parts of T and T', corresponding to the first i - 1 
 nodes traversed in preorder. Those parts must be isomorphic, otherwise we 
 could not have (x-j , .. . , x £ _ 1 )=(x* .... ,x^-,), by the equivalence (1)^(2). Hence, x<x' 
 iff all the leaves among the first I - 1 nodes (in preorder traversing) of 
 T and T' are in the same levels, correspondingly, but the leaf now scanned in 
 T (corresponding to x = 0) is in a lower level than that of the leaf that 
 will be next scanned in T' (this leaf will be a successor of the internal node 
 corresponding to x = 1 ) ; this happens iff the level sequence of T is less 
 than that of T' in Ruskey-Hu's ordering. 
 
10 
 
 (2) - (6) 
 
 i 
 
 x < x' iff for some i x. = x. for 1 = 1 , 2, .. . , £-1 and x - < 1 = x . 
 This happens iff while scanning both T and T 1 in preorder we meet internal 
 nodes and leaves correspondingly, but the £-th scanned node is a leaf b in 
 T and an internal node b' in T 1 . This happens iff while reading the pre- 
 order labeling of the vertices of T and T' in inorder , the two corresponding 
 permutations p and p' coincide until we read the labels of b and b'. At 
 this point in p we add the label of b as the next element, while in p' we add 
 a number greater than the label of b', but the labels of b and b' are the 
 same. This happens iff p<p' . ° 
 
 Corollary 4.9 : In order to generate B (lexicographically), it is sufficient 
 to generate all the above sequences x, y (lexicographically) or z(antilexico- 
 graphically) of Theorem 4.6. 
 
 Note : It should be clear how the conversion of a sequence into its correspond- 
 ing binary tree is made, and this fact is not discussed here. 
 4.10 : It is interesting to compare the feasibility condition in [RH] with the 
 
 following: 
 
 2n 
 Theorem 4.11 : For a 0,1-sequence x = {x-K the following are equivalent: 
 
 (1) x is a feasible sequence. 
 
 (2) Replacing the left most 100 pattern in x by 0, as long as 
 possible, terminates in (10) for some 1 < t < n. 
 
 (3) Erasing any 10 pattern in x, as long as possible, terminates 
 in the empty sequence. 
 
 Proof : Left to the reader. D 
 
 4.12 : The following algorithm generates the sequences lexicographically: 
 Algorithm 4.13 : (Generating the z sequences lexicographically): 
 
 Step 1 : Begin with z = {z. } = {1 , 2, . . . , n} 
 
 Step 2 : Scan the sequence from right to left. 
 
 

 11 
 
 Find the right most index j s.t. z. < 2. - 1. If no such 
 index exists - go to step (5). 
 Step 3 : The sequence z' = {zl} next to z is built as follows: 
 z. + z. for i < j 
 
 1 
 
 z. + z. + 1 
 
 J J 
 z i * z i_i + ] for i = j + 1, ... , n . 
 
 Let z ' be called z. 
 
 Step 4 : Go to step (2). 
 
 Step 5 : Exit. 
 Theorem 4.14 : Algorithm 4.13 generates all the feasible sequence z = (z-K 
 s.t. < z, < z 2 < .. . < z n and z. <_ 21 - 1 for i = 1 , 2, ... , n. 
 Proof : Trivial; the generating procedure in the algorithm follows the defini- 
 tion of lexicographic order. D 
 See Appendix for an example. 
 
12 
 
 V. THE RANKING FUNCTION 
 
 5.1 In this section we show how to determine the position Index(T) of 
 a given tree T e B in the lexicographic ordering of B . For this pur- 
 pose, we convert the tree to its corresponding z seqence as discussed in 
 Section 4, find the poisition index (z), of z in the lexicographic order- 
 ing of all these sequences, and set Index (T) = b> - index (z) + 1. 
 
 5.2 To determine the ranking function we define the doubly indexed sequence 
 
 a. . , « j * i-l : 
 
 1 >j 
 
 a 1.1-l ■ ' 
 
 for all i , 
 
 i,o ■ b i-i?rfi 1 i fora11 ' ■ 
 
 a. . = a. .,, + a. , . ■, otherwise . 
 
 
 
 1 1 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 1 
 
 1 
 
 
 
 2 
 
 1 
 
 5 
 
 3 
 
 1 
 
 14 
 
 9 
 
 4 
 
 1 
 
 42 
 
 28 
 
 14 
 
 5 
 
 1 
 
 132 
 
 90 
 
 48 
 
 20 
 
 6 
 
 1 
 
 429 
 
 297 
 
 165 
 
 75 
 
 27 
 
 7 
 
 1 
 
 1430 
 
 1001 
 
 572 
 
 275 
 
 110 
 
 35 
 
 8 
 
 1 
 
 The Sequence a- • 
 > »J 
 
 Figure 3 
 
13 
 
 n 
 5.3 Given a sequence z = {z-h, let £ = initCz) denote the largest i s.t. 
 
 n-1 
 
 z-j= i (note that z-j=l , so init(z) >_ 1). Let z = {z^ 1 be the sequence 
 
 built from z by deleting z and setting z. + z. for j < £, z. «-z.., - 2 for 
 
 ~ J J J Jt" i 
 
 j > £. The following theorem is the key to the ranking algorithm: 
 Theorem 5.4 : The following definition of the function index(z) assigns to a 
 feasible sequence z = {z.}? its position in the lexicographic ordering of all 
 these sequences: 
 
 
 [ 1 if £ = n 
 
 index(z) = / 
 
 \ a n,£ + index(z) if a j n 
 Proof : It is clear that 
 
 index(z) = [number of feasible sequences z 
 s.t. init(z) > £ + 1] + 
 [position of z among all those feasible 
 sequences z for which init(z) = £J. 
 
 We show that the first term in this sum equals a „ and that the second one 
 
 n n,£ 
 
 is index(z~) (it is clear that if £ = n then index(z)=l - see step 1 in 
 Algorithm 4.13). 
 
 . =0: a_ « = :-rr- \ n n )'' s the number of sequences that begin 
 
 - _L 
 
 'n,0 " n+1 
 
 with 1, which is all our sequences. For i = n-l: a , = 1 and there is 
 
 M n,n-l 
 
 really only one sequence z s.t. init(z) >_ i + 1 = n, namely {1, 2, ... , n}. 
 
 For any other i the sequences z with init(z) _> i + 1 are partitioned 
 
 into those for which init(z) > i + 1 and those for which init(z) = z + 1, and 
 
 this is taken case by the relation a „ = a „,, + a , „ , .. , .. . 
 
 J n s z n,£+l n-l,fc-l. We leave the rest 
 
 of the induction details to the reader. 
 
 Second term All the sequences z s.t. init(z) = £ corresponds to a q,1- 
 
 sequence x = 11 . . . Qo^. . . .The sequence z as described above is exactly the 
 
 £ 2n-£ 
 
14 
 
 sequence that we get from x by omitting the first 10 patterns from the left 
 (encircled above). So the position of z among all the sequences z with 
 init(z) = i is exactly index(z). D 
 
 Example 5.5 : Given the tree in Figure 2, we have: 
 
 index(l,2,3,4,8,10,ll,14) = a g 4 + index(l ,2,3,6,8,9,12) 
 index(l,2,3,6,8,9,12) = a 7 3 + index(T ,2,4,6,7,10) 
 index(l,2,4,6,7,10) = a 6 2 + index ( 1 > 2 »4,5,8) 
 index(l,2,4,5,8) = a 5 2 + index O > 2 > 3 > 6 ) 
 index(l,2,3,6) = a 4 3 + index(l,2,4) 
 
 index(l,2,4) = a 3 2 + index n.2) 
 
 index(l,2) = 1 
 
 so 
 
 index(l,2,3,4,8,10,ll,14) = a Q ^ + a ? ^ + a e ^ + a 5>2 + a 4>3 + a 3)2 + 
 
 1 = 110 + 75 + 48 + 14 + 1 + 1 + 1 = 250. 
 Thus Index(T)=b 8 - 250 + 1 = 1181. 
 
 5.6 The solution to the recurrence relation for a. • is 
 
 a = j±2_/2i-j\ 
 d i,j 2i-j li-j-lj 
 
 (See [RH] for an analytic approach, or apply the technique in [YY: problem 83] 
 
 for a combinatorial approach , which is also found in [WH] ) . 
 
15 
 
 VI. THE UNRANKING PROCEDURE 
 
 6.1 Given a number i, 1 <_ i £ b , we show in this section how to find the 
 regular binary tree T with n internal vertices s.t. Index(T) = i. 
 
 As explained in 5.1, it is enough if we know how to do it for a 
 z sequence. 
 
 6.2 The algorithm we are presenting follows immediately from the way we 
 calculated index(z) using Theorem 5.4, so we shall skip the proof of it here. 
 We first bring an example: 
 
 Example 6.3 : To find T e EL s.t. Index(T) = 1181, we note that it is sufficient 
 to find the appropriate z sequence for which index(Z) = b g - Index(T) + 1 = 
 250. By Theorem 5.4 we know that 
 
 250 = a Q + index(z) 
 
 o, x, 
 
 As a consequence from Section 5, we choose i s.t. 
 
 a 8,£ < 250 ^ a 8,£-l 
 Here we choose i = 4, and get 250 = 110 + index(z), or index(z) = 
 
 90, where z is a sequence corresponding to a tree T e B-,. Next we get 
 
 90 = a-, 3 + index(z), or index(z) = 15, and so on. 
 
 At the end we have 
 
 250 = a Q A + a 7 - + a c 9 + a c + a, ^ + a- + a , . 
 8,4 7,3 6,2 5,2 4,3 3,2 2,1 
 
 From this decomposition of 250, we reconstruct the sequence z: 
 
 The last 1 tells us that at the \jery end (of computing index(z)) 
 we had {1,2}. Then a 3 2 tells us that a step before we had the sequence 
 {l,2,t}, and that after omitting the 2 and setting t «- t - 2 we get {1,2}. 
 So we have {1,2,4}. Recurring in this manner, we get: 
 
 a^ 3 -y 1 ,2,3,6 
 
 a 5,2 * 1 ' 2 ' 4 ' 5 ' 8 
 
16 
 
 a 6,2 "* 1 » 2 ' 4 ' 6 ' 7 »" | 
 a ?j3 + 1,2,3,6,8,9,12 
 
 a 8)4 - z = {1,2,3,4,8,10,11,14} 
 
 and now we build the tree (see Figure 2, and compare this example with 
 
 Example 5.5). 
 
 6.4 We bring here the algorithm that converts a number to its corresponding 
 
 z sequence. We are given n and t, and look for a sequence z = {z.-}-| (0 < z, 
 
 < z« < ... < z , z. <_ 2i - 1 for i = 1, 2, ... , n) s.t. index (z) = t : 
 
 Algorithm 6.5 : 
 
 Step 1 : A -*- t, j «- n 
 
 Step 2 : Find £. s.t. a. < A <_ a. 
 
 J J)^ J » * .r - 1 
 
 Step 3 : A + t - a 
 
 J.A. 
 
 j «- j - 1 
 
 if A > then go to Step 2. 
 n 
 Step 4 : We have now t = 2a. 
 
 J=J 
 
 
 
 J»*. 
 
 Step 5 : Set z *■ {1,2, ... , j n } 
 
 m * j + ] ' s * £ m 
 Step 6 : Change z as follows 
 
 z i+l * z t + 2 
 
 z s* s 
 
 1 < s 
 
 i = m , m- 1 , 
 
 Step 7 : m ■«- m + 1 
 
 , s 
 
 if m <n then s + i and go to step 6, 
 — m 3 r 
 
 Step 8 : Exit. 
 
17 
 
 VII. K-ARY TREES 
 
 7.1 We generalize in this section the results about correspondence between 
 
 trees and integer sequences and the generating procedure, as discussed in 
 
 Section 4, to k-ary trees. The proofs in this section are omitted. 
 
 7.2 Given a tree T e T(k,n) we label its nodes as in the binary case, and 
 
 kn 
 get a 0,1-sequence f(T) = {x.}, = x and an integer sequence h(x) = (z.jl = z 
 
 (the sequence y is not discussed here). See Figure 5 for an example. 
 
 A sequence related to f(T) is associated with a planted planar tree in [KL], 
 
 using a different approach . 
 
 x = 11000001000100100000 
 z = 1,2,8,12,15 
 
 
 Labeling a k-ary Tree 
 
 Figure 4 
 
 A 0,1-sequence so obtained will be called a k -feasible sequence . It 
 
 will be characterized as having the following property, as is shown in Theorem 
 
 7.6. 
 
 kn 
 
 7.3 Let a = {a-}-, be a seqence consisting of n l's and (k-l)n 0's. We say 
 
 i 
 that a has the k-dominating property if in each initial subsequence {a,-}-!* 
 
 1 £ £ £ kn, the accumulated number of l's is at least \t-\ ([~tl means "the 
 
 smallest integer greater or equal to t"). In other words, if this subsequence 
 
 contains si's, then it contains at most (k-l)s 0's. Note that for k = 2, we 
 
 get the dominating property as defined in 4.1. 
 
 7.4 It will be of interest to mention the following generalization of the 
 
18 
 
 box-office problem: The generalized box-office problem *'- kn people are wait- 
 ing in line at a box-office, n of them have k-1 1 dollar bills and the 
 rest (k-l)n have k dollar bills (assuming that k dollar bills do exist). Each 
 ticket costs k - 1 dollars. When the box-office opens, there is no money in 
 the till. Each customer buys one ticket. In how many ways can they stand such 
 that none of them will have to wait for change ? . 
 
 Clearly, for k = 2 we get the previous box-office problem of 4.4. 
 Furthermore as a result of theorem 7.8, the answer to the generalized box- 
 office problem is ( k .] )n+1 ( k „ n ) . 
 
 7.5 The following theorem makes the desired connection between k-ary trees 
 and the corresponding integer sequences and is a generalization of theorem 4.6: 
 Theorem 7.6 : The following sets are in -1 correspondence with one another: 
 
 (1) All the regular k-ary trees with n internal vertices. 
 
 kn 
 
 (2) All the 0,1 -sequences with n Ts and (k-l)n 0's {x-}, , having 
 
 the k-dominating property. 
 
 n 
 
 (3) All the integer sequences {z-}-. s.t. < z, < z ? < ... < z and 
 
 z. <_ ki - (k-1) for i = 1, 2, ... , n. 
 7.7 The follwoing generalization of Theorem 4.8 is the basis for our generat- 
 ing discussion. The proof follows immediately from that of Theorem 4.8. Note 
 that in [TR1 , 2] the permutation associated with a tree is not immediately 
 extended from the binary case. Still, the order is the same. 
 Theorem 7.8 : Let T and T' be two regular k-ary trees with n internal nodes, and 
 let the corresponding x, z and x', z' sequences be as defined above. The 
 following are equivalent: 
 
 (1) T < T' (definition 2.8) 
 
 (2) x < x' 
 
 (3) z >z' 
 
 
 * 
 
 See [DM] for a related problem . 
 
19 
 
 (4) T < T' according to Ruskey-Hu's level sequence as extended to 
 k-ary trees ( see 3.1 ) . 
 
 (5) T < T' according to Trojanowski 's permutation as extended to 
 k-ary tree ( see 3.2). 
 
 Corollary 7.9 : In order to generate (lexicographically) all the regular k-ary 
 
 trees with n internal nodes, it is sufficient to generate all the sequences 
 
 x(lexicographically) or z(antilexicographically) of Theorem 7.6. 
 
 The note after corollary 4.9, and the discussion in 4.10 - 4.11 
 
 (replacing 100 and 10 patterns with 100 . . . y and 100 • --O, patterns respect- 
 
 k ~~k^T 
 ively), should be applied here as well. 
 
 Algorithm 7.10 : (Generating the z sequences lexicographically): Exactly 
 
 as algorithm 4.13, with one change. In step (2) replace the upper bound 2j - 1 
 
 by kj- (k-1). 
 
 7.11 All the discussion in Section 4, following algorithm 4.13, applies here. 
 
 7.12 : As for the ranking function and unranking procedure we were not able 
 
 to extend the algorithms from the binary case to the k-ary case; yet, as follows 
 
 from Theorem 7.8, we can use the appropriate algorithms in [TR 1, 2] for these 
 
 tasks. It should be noted that generating the k-ary trees by the z sequences, 
 
 is simpler than that of level numbers ([RH]) or permutations ([KM 1], [TR 1, 2], 
 
 [KN0]). 
 
 7.13 : Two methods for ranking and unranking k-ary trees- one of which is 
 
 presented among other combinatorial objects ([ZR]), and the other performs 
 
 these tasks in an order different from the two definitions, as discussed in 
 
 Section 2 ([ZA])- are a subject of future work. 
 
20 
 
 VIII. SUMMARY 
 
 8.1 After discussing the term "lexicographic order of trees," we have seen 
 algorithms that generate all the regular binary trees with n internal vertices 
 lexicographic order, find the position of a given tree in this ordering and 
 find a tree given its position. For regular k-ary trees we treated the 
 generating algorithm. 
 
 8.2 As explained in 2.2, there is a simple 1 - 1 correspondence between all 
 the regular k-ary trees with n internal vertices and all the k-ary trees with 
 n vertices, so the results mentioned in 8.1 can be extended to this new class. 
 
 8.3 If we take any tree with n edges and travese it in the way shown in 
 Figure 5, (see [BN]),when 1 means "going down" and means "going up" we get 
 a 0,1-sequence 
 
 110110001010 
 
 
 
 Figure 5 
 which have the dominating property - the number of "going down" must be at 
 least as that of "going up," otherwise we will get above the root! Consequently, 
 our algorithms can be used to generate, rank and unrank all the trees with n 
 edges. See Appendix for an example. 
 
 8.4 In finding the ranking function we defined the sequence a- • and then 
 proved that a is the number of feasible sequences z s.t.init(z) >_ n + 1 . We 
 could as well choose the alternative way: define a. . by this last property, 
 and then prove that a. . satisfies the relations by which we defined it. 
 

 21 
 
 8.5 The following remark concerns the difference between the two ordering of 
 trees (definitions 2.5 and 2.8): Given 2 trees T and T'^to determine whether 
 T < T, we scan both trees in preorder. Using definition 2.5, we compare the 
 sizes of the subtrees rooted at the concurrent nodes; in other words, we use 
 a global information concerning those nodes. On the other hand, using 
 definition 2.8 we compare the characters of the concurrent nodes (whether 
 they are internal nodes or leaves); in other words, we use local information 
 concerning those nodes. 
 
 ACKNOWLEDGE 
 
 I wish to thank Professor C. L. Liu for many helpful suggestions 
 
 while working on this paper. 
 

 22 
 
 APPENDIX 
 
 We show here all the 14 regular binary trees with 4 internal nodes 
 in lexicographic order. To each of them we assign the structures and sequences 
 introduced in the paper, as follows: 
 
 (1) The x sequence (0, 1-sequence) 
 
 (2) The y sequence 
 
 (3) The z sequence 
 
 (4) Ruskey and Hu's level sequence 
 
 (5) Trojanowski 's permutation 
 
 (note that reading only the internal vertices in Trojanowski 's 
 labeling in preorder (Root-Left-Right) gives us again the 
 corresponding z sequence, and the reasoning for this is left 
 to the reader. ) 
 
 (6) The corresponding binary tree with 4 vertices. 
 
 (7) The corresponding tree with 4 edges. 
 
 (8) The index while being ordered according to definition 2.5. 
 
 
23 
 
 
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26 
 
 REFERENCES 
 
 [BM] N. G. DeBruijn and B. J. M. Morselt, "A Note on Plane Trees," 
 Journal of Combinaotiral Theory 2, 27-34 (1967). 
 
 [DM] A. Dvoretzky and T. Motzkin, "A Problem of Arrangements," Duke 
 Math. Journal 14 (1947), 305-313. 
 
 [GA] M. Gardner, "Mathematical Games": Catalan Numbers, Scientific 
 American , June 1976, pp. 120-122. 
 
 
 [GN] B. N. Gnedenko, The Theory of Probability , Chelsea Publishing Company, 
 N.Y., N.Y., 1962. 
 
 [KL] D. A. Klarner, "Correspondences Between Plane Trees and Binary Sequences," 
 Journal of Combinatorial Theory 9, 401-411 (1970) 
 
 [KNO] Gary D. Knott, "A Numbering System for Binary Trees," Communications of 
 the ACM , Vol. 20, Number 2, February 1977. 
 
 [KN1] Donald E. Knuth, The Art of Computer Programming Vol. 1: Fundamental 
 Algorithms, Addi son-Wesley, Reading, MA 1968. 
 
 [KN3] Donald E. Knuth, The Art of Computer Programming Vol. 3: Sorting and 
 Searching, Addi son-Wesley, Reading, MA 1973. 
 
 [LI1] C. L. Liu, Topics in Combinatorial Mathematics Mathematical Association 
 of America, 1972. 
 
 [LI2] C. L. Liu, Elements of Discrete Mathematics McGraw-Hill, 1977. 
 
 [RH] F. Ruskey and T. C. Hu, "Generating Binary Trees Lexicographically," 
 SIAM Journal on Computing , to appear. 
 
 [TR1] Anthony E. Trojanowski, "On the Ordering, Enumeration and Ranking of 
 k-ary Trees," Tech. Report UIUCDCS-R-77-850, Department of Computer 
 Science, University of Illinois at Urbana-Champaign, February, 1977. 
 
 [TR2] Anthony E. Trojanowski, "Ranking and Listing Algorithms for k-ary Trees," 
 SIAM Journal on Computing , to appear. 
 
 [WH] W. A. Whitworth, "Arrangements of m Things of One Sort and m Things of 
 Another Sort Under Certain Conditions of Priority," Messenger of Math. 
 8 (1878), 105-114. 
 
 [YY] A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with 
 Elementary Solutions Vol. 1: Combinatorial Analysis and Probability 
 Theory | Hoi den-Day, 1964. 
 
 [ZA] S. Zaks, "Generating k-ary Trees Lexicographically," to appear. 
 
 [ZR] S. Zaks and D. Richards, "Generating Trees and Other Combinatorial Objects 
 Lexicographically," to appear. 
 
 
 
.LIOGRAPHIC DATA 
 EET 
 
 1. Report No. 
 
 UIUCDCS-R-77-888 
 
 2. 
 
 3. Recipient's Accession No. 
 
 "itle and Subtitle 
 
 5. Report Date 
 
 October, 1977 
 
 
 6. 
 
 uthor(s) 
 
 S. Zaks 
 
 8. Performing Organization Rept. 
 No. 
 
 "erforming Organization Name and Address 
 
 Department of Computer Science 
 
 10. Project/Task/Work Unit No. 
 
 University of Illinois at Urbana-Champaign 
 Urbana, IL 61801 
 
 11. Contract /Grant No. 
 
 MCS-73-03408 
 
 Sponsoring Organization Name and Address 
 
 National Science Foundation 
 
 13. Type of Report & Period 
 Covered 
 
 Washington, DC 
 
 14. 
 
 Supplementary Notes 
 
 Abstracts 
 
 We show a one-one correspondence between all the regular binary 
 trees with n internal nodes and certain integer sequences, an algorithm for 
 generating these trees lexicographically, and the ranking function and the 
 corresponding unranking procedure. We then extend the generating algorithm 
 to k-ary trees. Relations to existing algorithms are discussed. 
 
 Key Words and Document Analysis. 17a. Descriptors 
 
 k-ary tree, ordered tree, lexicographic order 
 
 , ranking, and 
 
 unranking 
 
 Identifiers/Open-Ended Terms 
 COSATI Field/Group 
 
 
 
 Availability Statement 
 
 19. Security Class (This 
 Report) 
 
 UNCLASSIFIED 
 
 21. No. of Pages 
 
 
 20. Security Class (This 
 Page 
 
 UNCLASSIFIED 
 
 22. Price 
 
 * NT IS- 35 (10-70) 
 
 USCOMM-DC 40329-P71 
 
J M 2 5 1978 
 
OCT 2" MO 
 
 W ^OVND ( 
 
UNIVERSITY OF ILLINOIS-URBAN A 
 510 84 IL6R no COO? no 886 893(1977 
 Generating binary trees lexicographical! 
 
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