Information Processing Letters 109 (2009) 805–810

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Information Processing Letters

www.elsevier.com/locate/ipl

(2,1)-Total number of trees with maximum degree three ✩

Weifan Wang ∗, Dong Chen

Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 14 October 2008Received in revised form 16 March 2009Accepted 30 March 2009Available online 1 April 2009Communicated by M. Yamashita

Keywords:Combinatorial problemsMaximum degree(2,1)-Total labellingTree

The (2,1)-total labelling number λt2(G) of a graph G is the width of the smallest range of

integers that suffices to label the vertices and the edges of G such that no two adjacentvertices have the same label, no two adjacent edges have the same label and the differencebetween the labels of a vertex and its incident edges is at least 2. It is known that everytree T with maximum degree � has �+ 1 � λt

2(T ) � �+ 2. In this paper, we characterizecompletely the (2,1)-total number of trees with � = 3.

© 2009 Elsevier B.V. All rights reserved.

1. Introduction

Motivated by the Frequency Channel Assignment prob-lem, Griggs and Yeh [5] introduced the L(2,1)-labelling ofgraphs. This notion was subsequently generalized to theL(p,q)-labelling problem of graphs. Let p and q be twononnegative integers. An L(p,q)-labelling of a graph G isa function f from its vertex set V (G) to the set {0,1,

. . . ,k} such that | f (x) − f (y)| � p if x and y are adja-cent, and | f (x) − f (y)| � q if x and y are at distance 2.The L(p,q)-labelling number λp,q(G) of G is the small-est integer k such that G has an L(p,q)-labelling f withmax{ f (v) | v ∈ V (G)} = k.

The L(p,q)-labelling of graphs have been studied ratherextensively in recent years [2,8,11,13,14].

Whittlesey, Georges and Mauro [15] investigated theL(2,1)-labelling of incidence graphs. The incidence graphof a graph G is the graph obtained from G by replacingeach edge by a path of length 2. The L(2,1)-labelling ofthe incidence graph of G is equivalent to an assignmentof integers to each element of V (G) ∪ E(G) such that ad-jacent vertices have different labels, adjacent edges have

✩ Research supported partially by NSFC (No. 10771197).

* Corresponding author.E-mail address: wwf@zjnu.cn (W. Wang).

0020-0190/$ – see front matter © 2009 Elsevier B.V. All rights reserved.doi:10.1016/j.ipl.2009.03.027

different labels, and incident vertex and edge have the dif-ference of labels by at least 2. This labelling is called (2,1)-total labelling of graphs, which was introduced by Havetand Yu [6] and generalized to the (d,1)-total labelling. Letd � 1 be an integer. A k-(d,1)-total labelling of a graph Gis a function f from V (G) ∪ E(G) to the set {0,1, . . . ,k}such that f (u) �= f (v) if u and v are two adjacent ver-tices, f (e) �= f (e′) if e and e′ are two adjacent edges, and| f (u)− f (e)| � d if a vertex u is incident to an edge e. The(d,1)-total number, denoted by λt

d(G), is the smallest inte-ger k such that G has a k-(d,1)-total labelling. When d = 1,the (1,1)-total labelling is the well-known total coloring ofa graph.

Let �(G) denote the maximum degree of a graph G . Incase of no confusion, we denote simply �(G) by �. The(d,1)-total labelling for a few special graphs have beenstudied, e.g., complete graphs and graphs with � � 5 [6],complete bipartite graphs [10], planar graphs [1], outerpla-nar graphs [3], products of graphs [4], graphs with a givenmaximum average degree [12], etc. A more general gener-alization of total coloring of graphs, so-called [r, s, t]-color-ing, was defined and investigated in [9].

For a tree T , it is easy to prove that � + 1 � λt2(T ) �

� + 2. In this paper, we give a complete characterizationfor a tree T with � = 3 to have λt (T ) = 4 or λt (T ) = 5.

2 2

806 W. Wang, D. Chen / Information Processing Letters 109 (2009) 805–810

Fig. 1. A bad tree T ∗ .

Fig. 2. Trees M and K1,3.

The paper is organized as follows. In Section 2, we es-tablish a structural property of trees under consideration.In Section 3, we give a sufficient and necessary conditionfor a tree T with � = 3 to have λt

2(T ) = 4 or λt2(T ) = 5. An

open problem on characterizing trees with � � 4 is pro-vided in Section 4.

2. A structural lemma

In order to obtain our main result in this paper, weneed to investigate the structural properties of trees withmaximum degree three. We start with some useful nota-tion.

For a tree T , we use |T | to denote the cardinality ofits vertex set, i.e., |T | = |V (T )|. Let dT (v) (or simply d(v))denote the degree of a vertex v in T . A vertex of de-gree k is called k-vertex. A vertex v is said to be major ifd(v) = �, minor if d(v) < �, a leaf if d(v) = 1, and a han-dle if d(v) > 1 and v is adjacent to at most one vertexof degree greater than one. A k-handle is a handle of de-gree k. Obviously, every tree T with |T | � 3 that is not astar contains at least two handles. For a path P , let �(P )

denote the length of P , i.e., the number of edges in P . Weuse V i(T ) to denote the set of i-vertices in T .

A path Pk+2 = x0x1 . . . xkxk+1 of length k + 1 is calleda k-chain if d(x0) > 2, d(xk+1) > 2, and d(xi) = 2 for alli = 1,2, . . . ,k. We note that a 0-chain is exactly an edgewith two ends of degree more than 2.

Suppose that a tree T with � = 3 has a 4-(2,1)-totallabelling f using the label set B = {0,1,2,3,4}. From f ,we can define a new 4-(2,1)-total labelling f ′ of T byf ′(x) = 4 − f (x) for each x ∈ V (T ) ∪ E(T ), where 0 is re-placed by 4; 1 by 3; 3 by 1; 4 by 0; whereas 2 keepsunchanged. We call f ′ the symmetric labelling of f .

Given a tree T with � = 3, a subtree T ∗ of T is calledbad if it satisfies the following conditions (1)–(4):

(1) V 3(T ∗) �= ∅;

(2) V 1(T ∗) ∪ V 3(T ∗) ⊆ V 3(T ) and V 2(T ∗) ⊆ V 2(T );(3) every vertex x ∈ V 1(T ∗) is connected to some vertex

y ∈ V 3(T ∗) by a 1-chain;(4) every two closest vertices x, y ∈ V 3(T ∗) are con-

nected by a 2-chain or 1-chain.

When V 3(T ∗) = {x}, T ∗ consists of three 1-chainsxv1 v2, xu1u2 and xw1 w2, where v2, u2, w2 are leaves ofT ∗ that are of degree 3 in T , and v1, u1, w1 are of degree2 in both T ∗ and T . A bad subtree T ∗ with |V 1(T ∗)| = 5,|V 2(T ∗)| = 9, and |V 3(T ∗)| = 3 is depicted in Fig. 1.

A tree T is called good if it contains neither bad sub-trees nor 3-vertex adjacent to two other 3-vertices. Twotrees M and K1,3, as shown in Fig. 2, are clearly good.

Lemma 1. Let T be a good tree that is neither M nor K1,3 . ThenT contains one of the following configurations (see Fig. 3):

(A1) a leaf u adjacent to a 2-vertex v;(A2) a 2-chain x0x1x2x3 such that x0 is a 3-handle;(A3) a 1-chain x0x1x2 such that x0 is a 3-handle and x2 is ad-

jacent to a 1-vertex or 3-vertex x′;(A4) a k-chain x0x1 . . . xkxk+1 with k � 3;(A5) a 3-vertex u lies in a 2-chain ux1x2x3 and two 1-chains

uy1 y2 and uz1z2 such that both y2 and z2 are 3-handles;(A6) a 3-vertex u is adjacent to a leaf y and a 3-handle z, and

u lies in a (k − 1)-chain ux1x2 . . . xk for 2 � k � 3.

Proof. If G contains (A1), we are done. Thus, assume thatG does not contain (A1). Since G is neither M nor K1,3,it is easy to inspect that |T | � 7. Let Q = t1t2 . . . tm bea longest path in T . Since |T | � 7 and (A1) is excludedfrom G , we see that m � 5, t1, tm are leaves, and t2, tm−1are 3-vertices. Let y denote the neighbor of t2 differentfrom t1 and t3. Then y must be a leaf, for otherwise Twill contain a path Q ′ with �(Q ′) > �(Q ), contradictingthe choice of Q . Since m � 5, it follows that d(t3) � 2. Theproof is divided into two cases as follows:

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Fig. 3. The configurations (A1)–(A6) in Lemma 1.

Case 1. d(t3) = 3.Let w �= t2, t4 denote the third neighbor of t3. Since

T does not contain a 3-vertex adjacent to two 3-vertices,d(w) �= 3. If d(w) = 2, let w ′ �= t3 denote the second neigh-bor of w . Since G does not contain (A1), w ′ is not a leaf.Let w ′′ denote a neighbor of w ′ different from w . ThenQ ′ = w ′′w ′wt3t4 . . . tm is a path in T with �(Q ′) > �(Q ),contradicting the choice of Q . Thus, d(w) = 1. Since Tdoes not contain a 3-vertex adjacent to two 3-vertices,d(t4) �= 3. Furthermore, d(t4) = 2 because T is not M . Since(A1) is excluded from T , d(t5) � 2. If d(t5) = 3, (A6) holdsfor the case k = 2. So suppose that d(t5) = 2. With the sim-ilar reason, d(t6) � 2. When d(t6) = 2, (A4) holds. Whend(t6) = 3, (A6) holds for the case k = 3.

Case 2. d(t3) = 2.It follows that d(t4) � 2 since G does not contain (A1).

If d(t4) = 2, then d(t5) � 2. When d(t5) = 2, (A4) holds.When d(t5) = 3, (A2) holds.

Suppose now that d(t4) = 3. Let s denote the neigh-bor of t4 different from t3 and t5. If d(s) = 1 or d(s) = 3,

then (A3) holds. So, assume that d(s) = 2. Let s′ denotethe neighbor of s different from t4. Since G does not con-tain (A1), d(s′) � 2. Let z denote an arbitrary neighbor ofs′ different from s. We claim that z is a leaf. Indeed, ifd(z) � 2, let z′ be a neighbor of z different from s′ . ThenQ ′ = z′zs′st4t5 . . . tm is a path with �(Q ′) > �(Q ), contra-dicting the choice of Q . Hence s′ is a handle. Since G doesnot contain (A1), d(s′) = 3.

If d(t5) = 1 or d(t5) = 3, then (A3) holds. Assume thatd(t5) = 2, which implies that d(t6) � 2. If d(t6) = 3, thenthe vertex subset {t2, t3, . . . , t6, s, s′} induces a bad sub-tree T ∗ of T , a contradiction. Thus, d(t6) = 2, implyingd(t7) � 2. If d(t7) = 2, (A4) holds; otherwise (A5) holds.This completes the proof of the lemma. �3. A characterization

It is known that every tree T with � = 3 has 4 �λt

2(T ) � 5. In this section, we give a complete character-ization for such tree T to have λt

2(T ) = 4 or λt2(T ) = 5. In

the proof of the main result, i.e., Theorem 6, we need thefollowing two known lemmas:

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Lemma 2. (See Havet and Yu, 2002 [6].)

(1) If H is a subgraph of G, then λt2(H) � λt

2(G).(2) If λt

2(G) = �+1 and f is a (�+1)-(2,1)-total labelling ofG using the labels 0,1, . . . ,� + 1, then every major vertexv of G has f (v) = 0 or f (v) = � + 1.

Lemma 3. (See Huang et al., 2008 [7].) If a graph G contains amajor vertex v adjacent to at least d(v)−1 other major vertices,then λt

2(G) � � + 2.

The following Lemmas 4 and 5 play an important rolein the proof of Theorem 6:

Lemma 4. Suppose that T is a tree with � = 3 and λt2(T ) =

4, and f is a 4-(2,1)-total labelling of T using the label setB = {0,1,2,3,4}. Let P = x0x1 . . . xmxm+1 be an m-chain ofT . Then the following statements hold:

(1) If m = 0, then f (x0x1) = 2.(2) If m = 1, then neither f (x0x1) nor f (x1x2) is 2.(3) If m = 2, then at most one of f (x0x1) and f (x2x3) is 2.

Proof. By definition, we see that d(x0) = d(xm+1) = 3and d(xi) = 2 for i = 1,2, . . . ,m. By Lemma 2(2), f (x0),

f (xm+1) ∈ {0,4}, say f (x0) = 0.(1) If m = 0, then P = x0x1. It is easy to see that

f (x1) = 4, and hence f (x0x1) = 2.(2) If m = 1, then P = x0x1x2. Assume to the contrary

that f (x0x1) = 2. Then we must have that f (x1) = 4, andhence f (x2) = 0. However, the edge x1x2 cannot be prop-erly labelled, a contradiction. Similarly, we can prove thatf (x1x2) �= 2.

(3) If m = 2, then P = x0x1x2x3. If f (x0x1) = f (x2x3) =2, then it is easy to derive that f (x1), f (x2) ∈ {0,4}, i.e.,f (x1) = 4 and f (x2) = 0. However, the edge x1x2 cannot belabelled properly. The proof of the lemma is complete. �Lemma 5. Let T be a tree with � = 3. Let u be a 3-handle ofT adjacent to two leaves v1, v2 and the other vertex v3 . If thesubtree T − {v1, v2} has a 4-(2,1)-total labelling f using B ={0,1,2,3,4} with f (u) ∈ {0,4}, then f can be extended to T .

Proof. Without loss of generality, assume that f (u) = 0.Thus, f (uv3) ∈ {2,3,4}. We label, respectively, uv1, uv2,

v1, v2 with 3,4,1,2 if f (uv3) = 2, with 2,4,4,2 iff (uv3) = 3, and with 2,3,4,1 if f (uv3) = 4. It is easyto inspect that the labelling obtained is a 4-(2,1)-total la-belling of T . The proof of the lemma is complete. �

Now, we state and prove our main result in this paper:

Theorem 6. If T is a tree with � = 3, then λt2(T ) = 4 if and

only if T is good.

Proof. Necessity. Assume that λt2(T ) = 4. Let f be a 4-

(2,1)-total labelling of T using B = {0,1,2,3,4}. If anedge e is labelled with 2, we call e a 2-edge. It sufficesto show that T is good, i.e., having neither bad subtrees

nor 3-vertex adjacent to two 3-vertices. The latter holdsobviously by Lemma 3.

Assume the contrary that T contains a bad subtreeT ∗ . Then |V 3(T ∗)| � 1 and T ∗ consists of 2-chains and1-chains. For any vertex v ∈ V 3(T ∗), f (v) ∈ {0,4} byLemma 2(2). If f (v) = 0, then 2, 3, 4 are used to color theedges incident to v . If f (v) = 4, then 0, 1, 2 are used tocolor the edges incident to v . Thus, there always exists anedge incident to v , denoted by ev , which is colored with 2.Note that ev lies in a 2-chain, denoted by P v , and every 2-chain contains at most one 2-edge by Lemma 4. It followsthat if u and w are two distinct 3-vertices in V 3(T ∗), thenthe corresponding 2-edges eu and ew , further 2-chainsPu and P w , are different (i.e., internally vertex-disjoint).Hence the number of 2-chains in T ∗ is at least |V 3(T ∗)|.However, since any two closest 3-vertices in V 3(T ∗) areconnected by a 2-chain or 1-chain in T ∗ , there exist atmost |V 3(T ∗)|−1 2-chains in T ∗ . This contradiction showsthat T cannot contain bad subtrees.

Sufficiency. It is straightforward to see that λt2(T ) � 4

for any tree T with �(T ) = 3. Conversely, we are goingto prove that T has a 4-(2,1)-total labelling by inductionon |T |. If |T | � 4, this holds obviously. Assume that T isa good tree with |T | � 5. If T is M , then a 4-(2,1)-totallabelling of M is easily constructed. Thus, assume that T isnot M . By Lemma 1, T contains one of the configurations(A1)–(A6). Hence we need to deal with every possible case.In what follows, we always use B = {0,1,2,3,4} to denotea set of five labels.

(A1) There is a leaf u adjacent to a 2-vertex v .Let T ′ = T − u. It is easy to see that T ′ is a good

tree with �(T ′) = 3. By the induction hypothesis, T ′ hasa 4-(2,1)-total labelling f using B. Since there exist atmost four forbidden labels for uv from f , and then againthere exist at most four forbidden labels for u after f isextended to uv , f can be extended into a 4-(2,1)-total la-belling of T .

(A2) There is a 2-chain x0x1x2x3 such that x0 is a 3-handle.

Let y1 and y2 be the neighbors of x0 different from x1.Let T ′ = T − {x0, x1, y1, y2}. Obviously, T ′ is a good treewith �(T ′) = 3, and hence has a 4-(2,1)-total labelling fusing B. Since dT ′(x3) = 3, f (x3) ∈ {0,4} by Lemma 2(2),say f (x3) = 0. Thus, f (x2x3) ∈ {2,3,4}. Erasing the la-bel of x2, we, respectively, label x2, x1x2, x1, x0x1, x0 with4,0,2,4,0 if f (x2x3) = 2, with 1,4,0,2,4 if f (x2x3) = 3,and 1,3,0,2,4 if f (x2x3) = 4. By Lemma 5, the current la-belling can be extended to the whole tree T .

(A3) There is a 1-chain x0x1x2 such that x0 is a 3-handleand x2 is adjacent to a 1-vertex or 3-vertex x′ .

Let y1 and y2 be the neighbors of x0 different fromx1. Let T ′ = T − {x0, y1, y2}. Then T ′ is a good tree with�(T ′) = 3, and hence has a 4-(2,1)-total labelling f us-ing B. Since dT ′ (x2) = 3, Lemma 2(2) asserts that f (x2) ∈{0,4}, assuming that f (x2) = 0. Note by definition thatdT ′ (x′) = 1 or 3. If dT ′(x′) = 3, then since x2 and x′ aretwo adjacent 3-vertices in T ′ , f (x2x′) = 2 by Lemma 4(1),implying that f (x1x2) �= 2. Assume that dT ′(x′) = 1. If

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f (x1x2) = 2, we first switch the labels of x1x2 and x2x′ andthen relabel x′ with a appropriate label. Thus, x1x2 alwaysgets a label b �= 2 in both cases. Obviously, b ∈ {3,4}. Eras-ing the label of x1, we label, respectively, x1, x0x1, x0 with1,4,0 if b = 3, and with 2,0,4 if b = 4. Lemma 5 guar-antees that the current labelling can be extended to thewhole tree T .

(A4) There is a k-chain x0x1 . . . xkxk+1 with k � 3.Let T ′ = T − {x2, x3, . . . , xk−1}. Since k � 3, T ′ consists

of two subtrees T1 and Tk , where Ti denotes the tree con-taining the vertex xi for i = 1,k. It is easy to see that Ti

is a good tree with �(Ti) = 3 and hence, by the induc-tion hypothesis, admits a 4-(2,1)-total labelling f i usingB. The combination of f1 with fk produces a 4-(2,1)-totallabelling f of T ′ using B. Since dT ′(x0) = dT ′ (xk+1) = 3,f (x0), f (xk+1) ∈ {0,4} by Lemma 2(2), say f (x0) = 0. Thus,f (x0x1) ∈ {2,3,4}. Erasing the labels of x1 and xk , we con-sider the following two cases, depending on the parityof k:

(A4.1) k � 3 is odd.Without loss of generality, assume that f (xk+1) = 0, so

f (xkxk+1) ∈ {2,3,4}. In fact, if f (xk+1) = 4, we may usethe symmetric labelling of f on Tk such that xk+1 is la-belled with 0. It suffices to handle the following subcasesby symmetry of x0 and xk+1.

(A4.1.1) f (x0x1) = 2.

• If f (xkxk+1) = 2, we first label x1 with 4. Then wealternately label x1x2, x2, x2x3, . . . , xk−1xk, xk with re-peated uses of the label sequence 1,3,0,4.

• If f (xkxk+1) = 3, we first label x1, x1x2, x2, xk−1xk, xkwith 4,0,2,4,1, respectively. Then we alternatelylabel x2x3, x3, x3x4, . . . , xk−2xk−1, xk−1 with repeateduses of the label sequence 4,1,3,0.

• If f (xkxk+1) = 4, we first label xk with 2. Then wealternately label x1, x1x2, x2, . . . , xk−1, xk−1xk with re-peated uses of the label sequence 4,1,3,0.

(A4.1.2) f (x0x1) = 3 and f (xkxk+1) �= 2.

• If f (xkxk+1) = 3, we take the symmetric labellingof f on Tk such that xk+1 is labelled with 4 andxkxk+1 with 1. We first label x1, x1x2, x2, xk−1xk, xkwith 1,4,2,0,3, respectively. Then we alternatelylabel x2x3, x3, x3x4, . . . , xk−2xk−1, xk−1 with repeateduses of the label sequence 0,3,1,4.

• If f (xkxk+1) = 4, we first label x1 with 1. Then wealternately label x1x2, x2, x2x3, . . . , xk−1xk , xk with re-peated uses of the label sequence 4,0,3,1.

(A4.1.3) f (x0x1) = f (xkxk+1) = 4.We take the symmetric labelling of f on Tk such that

xk+1 gets 4 and xkxk+1 gets 0. After labelling x1 with 2, wealternately label x1x2, x2, x2x3, . . . , xk−1xk, xk with repeateduses of the label sequence 0,4,1,3.

(A4.2) k � 4 is even.With the same reasoning, we assume that f (xk+1) = 0.

Thus f (xkxk+1) ∈ {2,3,4}.

(A4.2.1) f (x0x1) = 2.

• If f (xkxk+1) = 2, we take the symmetric labelling of fon Tk such that xk+1 gets 4 and xkxk+1 keeps 2. Afterlabelling xk−1, xk−1xk, xk with 2,4,0, respectively, wealternately label x1, x1x2, x2, . . . , xk−2, xk−2xk−1 withrepeated uses of the label sequence 4,1,3,0.

• If f (xkxk+1) = 3, we take the symmetric labelling off on Tk such that xk+1 gets 4 and xkxk+1 gets 1. Afterlabelling x1, x1x2, x2 with 4,0,3, respectively, we alter-nately label x2x3, x3, x3x4, . . . , xk−1xk, xk with repeateduses of the label sequence 1,4,0,3.

• If f (xkxk+1) = 4, we take the symmetric labelling off on Tk such that xk+1 gets 4 and xkxk+1 gets 0. Afterlabelling x1, x1x2, x2 with 4,1,3, respectively, we alter-nately label x2x3, x3, x3x4, . . . , xk−1xk, xk with repeateduses of the label sequence 0,4,1,3.

(A4.2.2) f (x0x1) = 3 and f (xkxk+1) �= 2.

• If f (xkxk+1) = 3, we take the symmetric labelling of fon Tk such that xk+1 gets 4 and xkxk+1 gets 1. Afterlabelling x1, x1x2, x2, x2x3 with 1,4,0,2, respectively,and then erasing the label of xkxk+1, we alternatelylabel x3, x3x4, x4, . . . , xk , xkxk+1 with repeated uses ofthe label sequence 4,0,3,1.

• If f (xkxk+1) = 4, we first label xk−2xk−1, xk−1, xk−1xk,

xk with 2,4,0,2, respectively. Erasing the label of x0x1,we alternately label x0x1, x1, x1x2, . . . , xk−3xk−2, xk−2with repeated uses of the label sequence 3,1,4,0.

(A4.2.3) f (x0x1) = f (xkxk+1) = 4.We first label xk−2xk−1, xk−1, xk−1xk, xk with 2,4,0,2,

respectively. Erasing the label of x0x1, we alternately labelx0x1, x1, x1x2, . . . , xk−3xk−2, xk−2 with repeated uses of thelabel sequence 4,1,3,0.

(A5) There is a 3-vertex u lying in a 2-chain ux1x2x3 andtwo 1-chains uy1 y2 and uz1z2 such that both y2 and z2are 3-handles.

Let u1, u2 be the neighbors of y2 different from y1,and w1, w2 the neighbors of z2 different from z1. LetT ′ = T −{u1, u2, w1, w2, y2, z2, x1}+ ux2. Then T ′ is a treewith �(T ′) = 3. Let us prove that T ′ is good. Clearly, sinceT does not contain a 3-vertex adjacent to two 3-vertices,so does T ′ by its construction. Suppose that T ′ containsa bad subtree T ∗ . If u /∈ V (T ∗), then T ∗ is obviously abad subtree of T , a contradiction. Thus, u ∈ V (T ∗), andfurthermore x2 ∈ V (T ∗). That is, u must be a leaf of T ∗ .Let T̄ denote the subtree of T induced by the vertex setV (T ∗) ∪ {y1, y2, z1, z2, x1}. It is easy to see that T̄ is a badsubtree of T with y2, z2 as leaves and u as a 3-vertex. Thisis also impossible.

Let f be a 4-(2,1)-total labelling of T ′ using B.Since dT ′(u) = dT ′ (x3) = 3, f (ux2) �= 2 and f (x2x3) �= 2by Lemma 4(2). Without loss of generality, suppose that

810 W. Wang, D. Chen / Information Processing Letters 109 (2009) 805–810

f (x3) = 0, so f (x2x3) ∈ {3,4}. In order to extend f to T ,by Lemma 5, we only need to consider the following cases:

If f (x2x3) = 3, we label, respectively, u, ux1, x1, x1x2, x2with 4,2,0,4,1; uy1, y1, y1 y2, y2 with 0,2,4,0; and uz1,

z1, z1z2, z2 with 1,3,0,4.If f (x2x3) = 4, we label, respectively, u, ux1, x1, x1x2, x2

with 0,2,4,0,2; uy1, y1, y1 y2, y2 with 4,2,0,4; and uz1,

z1, z1z2, z2 with 3,1,4,0.

(A6) There is a 3-vertex u adjacent to a leaf y and a 3-handle z, and u lies in a (k − 1)-chain ux1x2 . . . xk for 2 �k � 3.

Let z1, z2 �= u denote the third neighbors of z, and letT ′ = T − z1. Then T ′ is a good tree with �(T ′) = 3, andhence has a 4-(2,1)-total labelling f using B.

(A6.1) k = 2.Since d(x2) = 3, f (x2) ∈ {0,4}, say f (x2) = 0. By

Lemma 4(2), f (x1x2) �= 2, implying f (x1x2) ∈ {3,4}. We la-bel, respectively, x1, x1u, u, uz, z, uy, y with 1,4,0,2,4,3,1 if f (x1x2) = 3, and with 1,3,0,2,4,4,1 if f (x1x2) = 4.

(A6.2) k = 3.We again assume that f (x3) = 0. Thus, f (x2x3) ∈

{2,3,4}. We label, respectively, x2, x2x1, x1, x1u, u, uz, z,uy, y with 4,0,2,4,0,2,4,3,1 if f (x1x2) = 2, with 1,4,2,

0,4,2,0,1,3 if f (x1x2) = 3, and with 2,0,3,1,4,2,0,0,3if f (x1x2) = 4. By Lemma 5, the current labelling can beextended to the whole tree T . �4. Concluding remarks

In this paper, we give a complete characterization fora tree T with maximum degree three according to their(2,1)-total numbers. It is not known what is the corre-sponding characterization for trees T with � � 4.

Let D�(T ) denote the set of integers k for which thereexist two vertices of maximum degree of distance at k ina tree T . Huang et al. [7] proved that if T is a tree with� � 4 and at least one of 1 and 2 does not belong toD�(T ), then λt

2(T ) = � + 1. This result is best possible inthe sense that, for any fixed integer k � 3, there exist in-

finitely many trees T with � � 4 and k /∈ D�(T ) such thatλt

2(T ) = � + 2.Before concluding this paper, we put forward to the fol-

lowing problem:

Problem. Give a complete classification for trees T with� � 4 according to their (2,1)-total numbers.

Acknowledgements

The authors thank the referees for their valuable sug-gestions to improve this work.

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