Download CS235102 Data Structures - National Chi Nan University

Chapter 5 Trees: Outline
 Introduction
 Representation Of Trees








Binary Trees
Binary Tree Traversals
Additional Binary Tree Operations
Threaded Binary Trees
Heaps
Binary Search Trees
Selection Trees
Forests
Introduction (1/8)
 A tree structure means that the data are organized
so that items of information are related by branches
 Examples:
Introduction (2/8)
 Definition (recursively): A tree is a finite set of
one or more nodes such that
 There is a specially designated node called root.
 The remaining nodes are partitioned into n>=0 disjoint
set T1,…,Tn, where each of these sets is a tree.
T1,…,Tn are called the subtrees of the root.
 Every node in the tree is the root of some
subtree
Introduction (3/8)
 Some Terminology
 node: the item of information plus the branches to each
node.
 degree: the number of subtrees of a node
 degree of a tree: the maximum of the degree of the
nodes in the tree.
 terminal nodes (or leaf): nodes that have degree zero
 nonterminal nodes: nodes that don’t belong to terminal
nodes.
 children: the roots of the subtrees of a node X are the
children of X
 parent: X is the parent of its children.
Introduction (4/8)
 Some Terminology (cont’d)
 siblings: children of the same parent are said to be
siblings.
 Ancestors of a node: all the nodes along the path
from the root to that node.
 The level of a node: defined by letting the root be at
level one. If a node is at level l, then it children are at
level l+1.
 Height (or depth): the maximum level of any node in
the tree
 Example
Introduction (5/8)
A is the root node
Property: (# edges) = (#nodes) - 1
B is the parent of D and E
C is the sibling of B
D and E are the children of B
D, E, F, G, I are external nodes, or leaves
A, B, C, H are internal nodes
Level
The level of E is 3
The height (depth) of the tree is 4
A
1
The degree of node B is 2
The degree of the tree is 3
B
C
The ancestors of node I is A, C, H
2
The descendants of node C is F, G, H, I
H
D
E
F
G
I
3
4
Introduction (6/8)
 Representation Of Trees
 List Representation
 we can write of Figure 5.2 as a list in which each of the
subtrees is also a list
( A ( B ( E ( K, L ), F ), C ( G ), D ( H ( M ), I, J ) ) )
 The root comes first,
followed by a list of sub-trees
Introduction (7/8)
 Representation Of
Trees (cont’d)
 Left ChildRight Sibling
Representation
Introduction (8/8)
 Representation Of Trees (cont’d)
 Representation
As A Degree
Two Tree
Binary Trees (1/9)
 Binary trees are characterized by the fact that
any node can have at most two branches
 Definition (recursive):
 A binary tree is a finite set of nodes that is either
empty or consists of a root and two disjoint binary
trees called the left subtree and the right subtree
 Thus the left subtree and the right subtree are
A
distinguished
B
A
B
 Any tree can be transformed into binary tree
 by left child-right sibling representation
Binary Trees (2/9)
 The abstract data type of binary tree
Binary Trees (3/9)
 Two special kinds of binary trees:
(a) skewed tree, (b) complete binary tree
 The all leaf nodes of these trees are on two adjacent levels
Binary Trees (4/9)
 Properties of binary trees
 Lemma 5.1 [Maximum number of nodes]:
1. The maximum number of nodes on level i of a binary
tree is 2i-1, i 1.
2. The maximum number of nodes in a binary tree of
depth k is 2k-1, k1.
 Lemma 5.2 [Relation between number of leaf
nodes and degree-2 nodes]:
For any nonempty binary tree, T, if n0 is the number
of leaf nodes and n2 is the number of nodes of
degree 2, then n0 = n2 + 1.
 These lemmas allow us to define full and
complete binary trees
Binary Trees (5/9)
 Definition:
 A full binary tree of depth k is a binary tree of death k
having 2k-1 nodes, k  0.
 A binary tree with n nodes and depth k is complete iff its
nodes correspond to the nodes numbered from 1 to n in
the full binary tree of depth k.
 From Lemma 5.1, the
height of a complete
binary tree with n nodes
is log2(n+1)
Binary Trees (6/9)

Binary tree representations (using array)

Lemma 5.3: If a complete binary tree with n nodes
is represented sequentially, then for any node with
index i, 1  i  n, we have
1. parent(i) is at i /2 if i  1.
If i = 1, i is at the root and has no parent.
2. LeftChild(i) is at 2i if 2i  n.
If 2i  n, then i has no left child.
A 1
3. RightChild(i) is at 2i+1 if 2i+1  n.
If 2i +1  n, then i has no left child
B 2
[1] [2] [3] [4] [5] [6] [7]
C 3
A
B
C — D — E
D
Level 1
Level 2
Level 3
4
5
E
6
7
Binary Trees (7/9)
 Binary tree representations (using array)
 Waste spaces: in the worst case, a skewed tree of depth
k requires 2k-1 spaces. Of these, only k spaces will be
occupied
 Insertion or deletion
of nodes from the
middle of a tree
requires the
movement of
potentially many nodes
to reflect the change in
the level of these nodes
Binary Trees (8/9)
 Binary tree representations (using link)
Binary Trees (9/9)
 Binary tree representations (using link)
Binary Tree Traversals (1/9)
 How to traverse a tree or visit each node in the
tree exactly once?
 There are six possible combinations of traversal
LVR, LRV, VLR, VRL, RVL, RLV
 Adopt convention that we traverse left before
right, only 3 traversals remain
LVR (inorder), LRV (postorder), VLR (preorder)
left_child
L: moving left
data right_child
V
:
visiting
node
R: moving right
Binary Tree Traversals (2/9)
 Arithmetic Expression using binary tree
 inorder traversal (infix expression)
A/B*C*D+E
 preorder traversal (prefix expression)
+**/ABCDE
 postorder traversal
(postfix expression)
AB/C*D*E+
 level order traversal
+*E*D/CAB
Binary Tree Traversals (3/9)
 Inorder traversal (LVR) (recursive version)
output: A / B * C * D + E
ptr
L
V
R
Binary Tree Traversals (4/9)
 Preorder traversal (VLR) (recursive version)
output: + * * / A B C D E
V
L
R
Binary Tree Traversals (5/9)
 Postorder traversal (LRV) (recursive version)
output: A B / C * D * E +
L
R
V
Binary Tree Traversals (6/9)
 Iterative inorder traversal
 we use a stack to simulate recursion
5
4 11
8
3 14
2 17
1
A B
/ *C D
* E
+
L
V
R
output: A / B*C * D + E
node
Binary Tree Traversals (7/9)
 Analysis of inorder2 (Non-recursive Inorder
traversal)
 Let n be the number of nodes in the tree
 Time complexity: O(n)
 Every node of the tree is placed on and removed
from the stack exactly once
 Space complexity: O(n)
 equal to the depth of the tree which
(skewed tree is the worst case)
Binary Tree Traversals (8/9)
 Level-order traversal
 method:
 We visit the root first, then the root’s left child, followed by the
root’s right child.
 We continue in this manner, visiting the nodes at each new
level from the leftmost node to the rightmost nodes
 This traversal requires a queue to implement
Binary Tree Traversals (9/9)
 Level-order traversal (using queue)
output: + * E * D / C A B
1 17 3 14 4 11 5
2
*+ E *
D /
8
C A B
FIFO
ptr
Additional Binary Tree Operations (1/7)
 Copying Binary Trees
 we can modify the postorder traversal algorithm only
slightly to copy the binary tree
similar as
Program 5.3
Additional Binary Tree Operations (2/7)
 Testing Equality
 Binary trees are equivalent if they have the same
topology and the information in corresponding nodes
is identical
V
L
R
the same topology and data as Program 5.6
Additional Binary Tree Operations (3/7)
 Variables: x1, x2, …, xn can hold only of two
possible values, true or false
 Operators: (and), (or), ¬(not)
 Propositional Calculus Expression
 A variable is an expression
 If x and y are expressions, then ¬x, xy, xy are
expressions
 Parentheses can be used to alter the normal order of
evaluation (¬ >  > )
 Example: x1  (x2  ¬x3)
Additional Binary Tree Operations (4/7)
 Satisfiability problem:
 Is there an assignment to make an expression true?
 Solution for the Example x1  (x2  ¬x3) :
 If x1 and x3 are false and x2 is true
 false  (true  ¬false) = false  true = true
 For n value of an expression, there are 2n
possible combinations of true and false
Additional Binary Tree Operations (5/7)
(x1  ¬x2)  (¬ x1  x3)  ¬x3

postorder traversal

data



value
X1

X3

X2
X3
X1
Additional Binary Tree Operations (6/7)
 node structure
 For the purpose of our evaluation algorithm, we
assume each node has four fields:
 We define this node structure in C as:
Additional Binary Tree Operations (7/7)
 Satisfiability function
 To evaluate the tree is
easily obtained by
modifying the original
recursive postorder
traversal
TRUE
node
TRUE
FALSE
FALSE
TRUE
TRUE
T
TRUE
TRUE
FALSE
F
F
FALSE FALSE
T
TRUE
T
L
R
V
Threaded Binary Trees (1/10)
 Threads
 Do you find any drawback of the above tree?
 Too many null pointers in current representation of
binary trees
n: number of nodes
number of non-null links: n-1
total links: 2n
null links: 2n-(n-1) = n+1
 Solution: replace these null pointers with some useful
“threads”
Threaded Binary Trees (2/10)
 Rules for constructing the threads
 If ptr->left_child is null,
replace it with a pointer to the node that would be
visited before ptr in an inorder traversal
 If ptr->right_child is null,
replace it with a pointer to the node that would be
visited after ptr in an inorder traversal
Threaded Binary Trees (3/10)
 A Threaded Binary Tree
t: true  thread
f: false  child
root
A
dangling
f B f
dangling
C
t E t
D
F
G
inorder traversal:
H
I
H D I B E A F C G
Threaded Binary
Trees (4/10)
 Two additional fields of the node structure,
left-thread and right-thread
 If ptr->left-thread=TRUE,
then ptr->left-child contains a thread;
 Otherwise it contains a pointer to the left child.
 Similarly for the right-thread
Threaded Binary Trees (5/10)
 If we don’t want the left pointer of H and the right
pointer of G to be dangling pointers, we may
create root node and assign them pointing to the
root node
Threaded Binary Trees (6/10)

Inorder traversal of a threaded binary tree

By using of threads we can perform an inorder
traversal without making use of a stack (simplifying
the task)
 Now, we can follow the thread of any node, ptr, to
the “next” node of inorder traversal
1. If ptr->right_thread = TRUE, the inorder successor
of ptr is ptr->right_child by definition of the threads
2. Otherwise we obtain the inorder successor of ptr by
following a path of left-child links from the right-child
of ptr until we reach a node with left_thread = TRUE
Threaded Binary Trees (7/10)
 Finding the inorder successor (next node) of a node
threaded_pointer insucc(threaded_pointer tree){
threaded_pointer temp;
temp = tree->right_child;
if (!tree->right_thread)
while (!temp->left_thread)
temp = temp->left_child;
return temp;
}
Inorder
tree
temp
Threaded Binary Trees (8/10)
 Inorder traversal of a threaded binary tree
void tinorder(threaded_pointer tree){
/* traverse the threaded binary tree inorder */
threaded_pointer temp = tree;
output: H D I B E A FC G
for (;;) {
temp = insucc(temp);
if (temp==tree)
tree
break;
printf(“%3c”,temp->data);
}
}
Time Complexity: O(n)
Threaded Binary Trees (9/10)

Inserting A Node Into A Threaded Binary Tree
 Insert child as the right child of node parent
1. change parent->right_thread to FALSE
2. set child->left_thread and child->right_thread to
TRUE
3. set child->left_child to point to parent
4. set child->right_child to parent->right_child
5. change parent->right_child to point to child
Threaded Binary Trees (10/10)
 Right insertion in a threaded binary tree
void insert_right(thread_pointer parent, threaded_pointer child){
/* insert child as the right child of parent in a threaded binary tree */
threaded_pointer temp;
root
child->right_child = parent->right_child;
parent
child->right_thread = parent->right_thread;
A
child->left_child = parent;
B
child->left_thread = TRUE;
X
C
child
parent->right_child = child;
parent->right_thread = FALSE;
temp
If(!child->right_thread){
parent
A
temp = insucc(child);
child
B
temp->left_child = child;
X
}
C
}
D
First Case
Second
Case
E
successor
F
Heaps (1/6)
 The heap abstract data type
 Definition: A max(min) tree is a tree in which the key
value in each node is no smaller (larger) than the key
values in its children. A max (min) heap is a complete
binary tree that is also a max (min) tree
 Basic Operations:
 creation of an empty heap
 insertion of a new elemrnt into a heap
 deletion of the largest element from the heap
Heaps (2/6)
 The examples of max heaps and min heaps
 Property: The root of max heap (min heap) contains
the largest (smallest) element
Heaps (3/6)
 Abstract data type of Max Heap
Heaps (4/6)
 Queue in Chapter 3: FIFO
 Priority queues




Heaps are frequently used to implement priority queues
delete the element with highest (lowest) priority
insert the element with arbitrary priority
Heaps is the only way to implement priority queue
machine service:
amount of time
(min heap)
amount of payment
(max heap)
factory:
time tag
Heaps (5/6)
 Insertion Into A Max Heap
 Analysis of insert_max_heap
 The complexity of the insertion function is O(log2 n)
insert 2
51
*n= 6
5
i= 1
6
7
3
[1]
20
21
[2]
15
[4]
parent sink
item upheap
[3]
[5]
14 10
20
52
[6]
[7]
2
5
 Deletion from a max heap
Heaps (6/6)
 After deletion, the
heap is still a
complete binary tree
 Analysis of
delete_max_heap
 The complexity of the
insertion function
is O(log2 n)
parent = 4
1
2
child = 8
2 [1]
4
[2]
<
15
20
[3]
15
14
[4]
*n= 5
4
[5]
14
10 10
2
item.key = 20
temp.key = 10
Binary Search Trees (1/8)
 Why do binary search trees need?
 Heap is not suited for applications in which arbitrary
elements are to be deleted from the element list
 a min (max) element is deleted
O(log2n)
 deletion of an arbitrary element
O(n)
 search for an arbitrary element
O(n)
 Definition of binary search tree:
 Every element has a unique key
 The keys in a nonempty left subtree (right subtree) are
smaller (larger) than the key in the root of subtree
 The left and right subtrees are also binary search trees
Binary Search Trees (2/8)
 Example: (b) and (c) are binary search trees
medium
smaller
larger
Binary Search Trees (3/8)
 Search:
Search(25) Search(76)
44
17
88
65
32
28
97
54
29
82
76
80
Binary Search Trees (4/8)
 Searching a
binary search
tree
O(h)
Binary Search Trees (5/8)
 Inserting into a binary search tree
An empty tree
Binary Search Trees (6/8)
 Deletion from a binary search tree
 Three cases should be considered
 case 1. leaf  delete
 case 2.
one child  delete and change the pointer to this child
 case 3. two child  either the smallest element in the right
subtree or the largest element in the left subtree
Binary Search Trees (7/8)
 Height of a binary search tree
 The height of a binary search tree with n elements
can become as large as n.
 It can be shown that when insertions and deletions
are made at random, the height of the binary search
tree is O(log2n) on the average.
 Search trees with a worst-case height of O(log2n) are
called balance search trees
Binary Search Trees (8/8)
 Time Complexity
 Searching, insertion, removal
 O(h), where h is the height of the tree
 Worst case - skewed binary tree
 O(n), where n is the # of internal nodes
 Prevent worst case
 rebalancing scheme
 AVL, 2-3, and Red-black tree
Selection Trees (1/6)
 Problem:
 suppose we have k order sequences, called runs, that
are to be merged into a single ordered sequence
 Solution:
 straightforward : k-1 comparison
 selection tree : log2k+1
 There are two kinds of selection trees:
winner trees and loser trees
Selection Trees (2/6)
 Definition: (Winner tree)
 a selection tree is the binary tree where each node
represents the smaller of its two children
 root node is the smallest node in the tree
 a winner is the record with smaller key
 Rules:
 tournament : between sibling nodes
 put X in the parent node  X tree
where X = winner or loser
 Winner Tree
Selection Trees (3/6)
sequential allocation
scheme
(complete
binary tree)
Each node represents
the smaller of its two
children
ordered sequence
Selection Trees (4/6)
 Analysis of merging runs using winner trees




# of levels: log2K  +1  restructure time: O(log2K)
merge time: O(nlog2K)
setup time: O(K)
merge time: O(nlog2K)
 Slight modification: tree of loser
 consider the parent node only (vs. sibling nodes)
Selection Trees (5/6)
 After one record has been output
6
6
6
6
15
Selection Trees (6/6)
 Tree of losers can be conducted by Winner tree
0
6
8
9
10
17
20
9
90
Forests (1/4)
 Definition:
 A forest is a set of n  0 disjoint trees
 Transforming a forest into a binary tree
 Definition: If T1,…,Tn is a forest of trees, then the
binary tree corresponding to this forest, denoted by
B(T1,…,Tn ):
 is empty, if n = 0
 has root equal to root(T1); has left subtree equal to
B(T11,T12,…,T1m); and has right subtree equal to
B(T2,T3,…,Tn)
 where T11,T12,…,T1m are the subtrees of root (T1)
Forests (2/4)
 Rotate the tree clockwise by 45 degrees
A
Leftmost child
A
B
C
G
E
D
Right sibling
F
H
B
I
E
F
C
D
G
H
I
 Forest traversals
Forests (3/4)
 Forest preorder traversal
(1)If F is empty, then return.
(2)Visit the root of the first tree of F.
(3)Traverse the subtrees of the first tree in tree preorder.
(4)Traverse the remaining tree of F in preorder.
 Forest inorder traversal
(1)If F is empty, then return
(2)Traverse the subtrees of the first tree in tree inorder
(3)Visit the root of the first tree of F
(4)Traverse the remaining tree of F in inorder
 Forest postorder traversal
(1)If F is empty, then return
(2)Traverse the subtrees of the first tree in tree postorder
(3)Traverse the remaining tree of F in postorder
(4)Visit the root of the first tree of F
preorder: A B C D E F G H I
inorder: B C A E D G H F I
A
E, D, G, H, F, I
B, C
preorder: A B C (D E F G H I)
inorder: B C A (E D G H F I)
A
D
B
C E
F, G, H, I
Forests (4/4)
Set Representation(1/13)
 S1={0, 6, 7, 8}, S2={1, 4, 9}, S3={2, 3, 5}
6
7
2
4
0
8
1
Si  Sj = 
9
3
5
 Two operations considered here
 Disjoint set union S1  S2={0,6,7,8,1,4,9}
 Find(i): Find the set containing the element i.
3  S3, 8  S1
Set Representation(2/13)
 Union and Find Operations
Make one of trees a subtree of the other
0
6
7
8
0
4
1
9
6
7
1
8
Possible representation for S1 union S2
4
9
Set Representation(3/13)
set
name pointer
S1
0
6
7
S2
8
4
S3
1
9
2
3
5
*Figure 5.41:Data Representation of S1S2and S3 (p.240)
Set Representation(4/13)
 Array Representation for Set
i
[0]
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
parent
-4
4
-3
2
-3
2
0
0
0
4
int find1(int i) {
for(; parent[i] >= 0; i = parent[i]);
return i;
}
void union1(int i, int j) {
parent[i] = j;
}
Program 5.18: Initial attempt at union-find function (p.241)
Set Representation(5/13)
union operation
O(n) n-1
n-1
find operation
O(n2) n
n-2
i
i 2



union(0,1), find(0)
union(1,2), find(0)
.
.
.
union(n-2,n-1),find(0)
0
degenerate tree
*Figure 5.43:Degenerate tree (p.242)
Set Representation(6/13)
weighting rule for union(i,j): if # of nodes in i < # in j then j the parent of i
Set Representation(7/13)
 Modified Union Operation
void union2(int i, int j)
Keep a count in the root of tree
{
int temp = parent[i] + parent[j];
if (parent[i] > parent[j]) {
parent[i] = j; /*make j the new root*/
parent[j] = temp;
}
else {
parent[j] = i;/* make i the new root*/
parent[i] = temp;
}
If the number of nodes in tree i is
}
less than the number in tree j, then
make j the parent of i; otherwise
make i the parent of j.
Set Representation(8/13)
Figure 5.45:Trees achieving worst case bound (p.245)
 log28+1
Set Representation(9/13)
 The definition of Ackermann’s function used
here is :
A ( p, q) =
P=0
2 q
0
q=0 and p >= 1


P>=1 and p = 1
0


 A( p  1, A( p, q  1))
p>=1 and q >= 2
Set Representation(10/13)
 Modified Find(i) Operation
Int find2(int i) {
int root, trail, lead;
for (root=i;parent[root]>=0;oot=parent[root])
;
for (trail=i; trail!=root; trail=lead) {
lead = parent[trail];
parent[trail]= root;
}
If j is a node on the path from
return root;
i to its root then make j a child
}
of the root
Set Representation(11/13)
0
0
1
4
2
3
5
1
6
2
4
3
5
6
7
7
find(7) find(7) find(7) find(7) find(7) find(7) find(7) find(7)
go up
reset
3
1
1
2
12 moves (vs. 24 moves)
1
1
1
1
1
Set Representation(12/13)
 Applications
 Find equivalence class i  j
 Find Si and Sj such that i  Si and j  Sj
(two finds)
 Si = Sj
do nothing
 Si  Sj
union(Si , Sj)
 example
0  4, 3  1, 6  10, 8  9, 7  4, 6  8,
3  5, 2  11, 11  0
{0, 2, 4, 7, 11}, {1, 3, 5}, {6, 8, 9, 10}
Set Representation(13/13)
Counting Binary trees(1/10)
 Distinct Binary Trees :
 If n=0 or n=1, there is only one binary tree.
 If n=2 and n=3,
Counting Binary trees(2/10)
 Stack Permutations
preorder:
inorder:
A
ABCDEF GHI
BCAED GH FI
A
D, E, F, G, H, I
B, C
A
C
D, E, F, G, H, I
B
C
D
B
E
F
G
I
H
Counting Binary trees(3/10)
 Figure5.49(c) with the node numbering
of Figure 5.50.Its preorder permutation
is 1,2…,9, and its inorder
1
permutation is 2,3,1,5,4,
4
7,8,6,9.
2
3
5
6
7
9
8
Counting Binary trees(4/10)
 If we start with the numbers1,2,3, then the
possible permutations obtainable by a stack are:
(1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,2,1)
 Obtaining(3,1,2) is impossible.
1
1
1
2
2
3
3
1
1
2
2
2
3
3
3
Counting Binary trees(5/10)
 Matrix Multiplication
 Suppose that we wish to compute the product of n matrices:
M1 * M2 . . .* Mn
 Since matrix multiplication is associative, we can perform
these multiplications in any order. We would like to know
how many different ways we can perform these
multiplications . For example, If n =3, there are two
possibilities:
(M1*M2)*M3
M1*(M2*M3)
Counting Binary trees(6/10)
 Let bn be the number of different ways to compute the
product of n matrices. Then b2  1, , b3  2 and b4  5 .
 Let M ij , i  j be the product M i * M i1 * ... * M j . .
The product we wish to compute is M ln by computing
any one of the products M 1i * M i 1,n ,1  i  n.
The number of distinct ways to obtain M 1i andM i 1,n
are bi and bni ,respectively. Therefore, letting bi =1,
we have:
n 1
bn 
b b
i 0
i
n i
,n 1
Counting Binary trees(7/10)
 Now instead let bn be the number of distinct binary
trees with n nodes. Again an expression for bn in
terms of n is what we want. Than we see that bn is the
sum of all the possible binary trees formed in the
following way: a root and two subtrees with bi and bn i 1
nodes, for 0  i  n . This explanation says that
bn 
n 1
b b
i 0
i
n i 1
,n  1
and
b0  1
bn
bi
bn-i-1
Counting Binary trees(8/10)
 Number of Distinct Binary Trees:
 To obtain number of distinct binary trees with n
nodes, we must solve the recurrence of Eq.(5.5).To
i
B
(
x
)

b
x
begin we let:
(5.6)
i0 i
 Which is the generating function for the number of
binary trees. Next observe that by the recurrence
relation we get the identity: xB2 ( x )  b( x )  1
 Using the formula to solve quadratics and the fact
(Eq. (5.5)) that B(0) = b0 = 1 ,we get
1  1  4x
B( x ) 
2x
Counting Binary trees(9/10)
 Number of Distinct Binary Trees:
 We can use the binomial theorem to expand
1  4 x 1 / 2 to obtain:
n


1 / 2
1
B( x )  1    ( 4 x ) 
2  n 0  n 


 1 / 2  m 2 m1 m
 
( 1) 2 x

m0 m  1
(5.7)
Counting Binary trees(10/10)
 Number of Distinct Binary Trees:
 Comparing Eqs.(5.6) and (5.7) we see that bn ,
which is the coeffcient of x n in B(x), is :  1 / 2 
  1n 2 2n1
n  1


Some simplification yields the more compact form
1  2n 
bn 
n  1  n 
which is approximately
bn  O(4
n
n
3/ 2
)