# Chapter 4 Binary Trees

A binary tree is an important type of structure which occurs very often. It is characterized by the fact that any node can have at most two branches, i.e.,there is no node with degree greater than two. For binary trees we distinguish between the subtree on the left and on the right, whereas for trees the order of the subtreewas irrelevant. Also a binary tree may have zero nodes. Thus a binary tree is really a different object than a tree.

Definition: A binary tree is a finite set of nodes which is either empty or consists of a root and two disjoint binary trees called the left subtree and the right subtree.

We can define the data structure binary tree as follows:

structure BTREE
declare CREATE( ) --> btree
ISMTBT(btree,item,btree) --> boolean
MAKEBT(btree,item,btree) --> btree
LCHILD(btree) --> btree
DATA(btree) --> item
RCHILD(btree) --> btree
for all
p,r in btree, d in item let
ISMTBT(CREATE)::=true
ISMTBT(MAKEBT(p,d,r))::=false
LCHILD(MAKEBT(p,d,r))::=p; LCHILD(CREATE)::=error
DATA(MAKEBT(p,d,r))::d; DATA(CREATE)::=error
RCHILD(MAKEBT(p,d,r))::=r; RCHILD(CREATE)::=error
end
end
BTREE

This set of axioms defines only a minimal set of operations on binary trees. Other operations can usually be built in terms of these. The distinctions between a binary tree and a tree should be analyzed. First of all there is no tree having zero nodes, but there is an empty binary tree. The two binary trees below are different. The first one has an empty right subtree while the second has an empty left subtree. If these are regarded as trees, then they are the same despite the fact that they are drawn slightly differently. ## Binary Tree Representations

A full binary tree of depth k is a binary tree of depth k having pow(2,k)-1 nodes. This is the maximum number of the nodes such a binary tree can have. A very elegant sequential representation for such binary trees results from sequentially numbering the nodes, starting with nodes on level 1, then those on level 2 and so on. Nodes on any level are numbered from left to right. This numbering scheme gives us the definition of a complete binary tree. A binary tree with n nodes and a depth k is complete iff its nodes correspond to the nodes which are numbered one to n in the full binary tree of depth k. The nodes may now be stored in a one dimensional array tree, with the node numbered i being stored in tree[i].

Lemma 5.3: If a complete binary tree with n nodes (i.e., depth=[LOG2N]+1) is represented sequentially as above then for any node with index i, 1 (i) parent(i) is at [i/2] if is not equal to 1. When i=1, i is the root and has no parent.
(ii) lchild(i) is at 2i if 2in, then i has no left child.
(iii) rchild(i) is at 2i+1 if 2i+1n, then i has no right child.

Proof: We prove (ii). (iii) is an immediate consequence of (ii) and the numbering of nodes on the same level from left to right. (i) follows from (ii) and (iii). We prove (ii) by induction on i. For i=1, clearly the left child is at 2 unless 2>n in which case 1 has no left child. Now assume that for all j, 1n in which case i+1 has no left child. This representation can clearly be used for all binary trees though in most cases there will be a lot of unutilized space. For complete binary trees the representation is ideal as no space is wasted. In the worst case a skewed tree of k will require pow(2,k)-1 spaces. Of these only k will be occupied.

While the above representation appears to be good for complete binary trees it is wasteful for many other binary trees. In addition, the representation suffers from the general inadequacies of sequential representations. Insertion or deletion of nodes from the middle of a tree requires the movement of potentially many nodes to reflect the change in level number of these nodes. These problems can be easily overcome through the use of a linked representation. Each node will have three fields leftchild, data and rightchild and is defined in Pascal as

type treepointer = ^treerecord;
treerecord = record
leftchild : treepointer;
data : char;
rightchild : treepointer;
end;

## Binary Tree Traversal

There are many operations that we often want to perform on trees. One notion that arises frequently is the idea of traversing a tree or visiting each node in the three exactly once. A full traversal produces a linear order for the information in a tree. This linear order may be familiar and useful. When traversing a binary tree we want treat each node and its subtrees in the same fashion. If we let L, D, R stand for moving left, printing the data, and moving right when at a node then there are six possible combinations of traversal: LDR, LRD, DLR, DRL, RDL, and RLD. If we adopt the convention that we traverse left before right then only three traversals remain: LDR, LRD, and DLR. To these we assign the names inorder, postorder and preorder because there is a natural correspondence between these traversals and producing the infix, postfix and prefix forms of an expression. Inorder Traversal: informally this calls for moving down the tree towards the left untilyou can go no farther. Then you "visit" the node, move one node to the right and continue again. If you cannot move to the right, go back one more node. A precise way of describing this traversal is to write it as a recursive procedure.
procedure inorder(currentnode:treepointer);
{currentnode is a pointer to a noder in a binary tree. For full
tree traversal, pass inorder the pointer to the top of the tree}
begin {inorder}
if currentnode <> nil
then
begin
inorder(currentnode^.leftchild);
write(currentnode^.data);
inorder(currentnode^.rightchild);
end
end; {of inorder}

Recursion is an elegant device for describing this traversal. A second form of traversal is preorder:
procedure preorder(currentnode:treepointer);
{currentnode is a pointer to a node in a binary tree. For full
tree traversal, pass preorder the ponter to the top of the tree}
begin {preorder}
if currentnode <> nil
then
begin
write(currentnode^.data);
preorder(currentnode^.leftchild);
preorder(currentnode^.rightchild);
end {of if}
end; {of preorder}

In words we would say "visit a node, traverse left and continue again. When you cannot continue, move right and begin again or move back until you can move right and resume. At this point it should be easy to guess the next thraversal method which is called postorder:

procedure postorder(currentnode:treepointer);
{currentnode is a pointer to a node in a binary tree. For full
tree traversal, pass postorder the pointer to the top of the tree}
begin {postorder}
if currentnode<> nil
then
begin
postorder(currentnode^.leftchild);
postorder(currentnode^.rightchild);
write(currentnode^.data);
end {of if}
end; {of postorder} To understand the subject better let's look at an example.    