Binary Trees (1) - York · PDF file8 Assume true for a tree with n nodes: E = I + 1. Now, we want to extend the tree by adding new nodes: 1. Cannot add only one node. •add one internal

1

Binary Trees (1)

Outline and Required Reading:

CSE 2011, Winter 2017, Section ZInstructor: N. Vlajic

• Binary Trees (§ 8.2)

2Binary Tree

Binary Tree – ordered tree with the following properties• each internal node has at most two children• children of internal nodes form ordered pairs:

left node – 1st, right node – 2nd

A

B C

D E F G

I J

1st childof A

2nd childof A

Application – representation of arithmetic expression, decision treeswith yes / no answers , …

3Binary Tree (cont.)

Example 1 [ binary tree for ‘yes/no’ decision process ]

Part of a decision tree containing yes-no questions that diagnose a problem with a TV …

4Properties of Proper Binary Trees

A

B C

E

Binary Tree Notation• n – number of nodes

• e – number of external nodes

• i – number of internal nodes

• h – height of the tree

• level – set of nodes with the same depth

Property 1.1 Level d has at most 2d nodes. (Applies to any binary tree.)

Proof Let us annotate max number of nodes at level d with mn(d).

Clearly, mn(0) = 1 = 20,mn(1) = 2 = 21,mn(2) = 4 = 22,mn(d) = 2*mn(d-1) for ∀d≥0 .

Hence, mn(d) = 2*mn(d-1) = 2*2*mn(d-2) = 2*[2*[..2*mn(0)]] = 2d

D

I J

F G

level 1

level 2

level 3

level 0

5

Property 1.2 A full binary tree of height h has (2h+1 – 1) nodes.

Proof

n = mn(0) + mn(1) + .. + mn(d=h) =

= 20 + 21 + 22 + … + 2h =

=

Full binary tree.

11−=

− ++

1h1h

21-2

2

Properties of Proper Binary Trees (cont.)

6

Property 2 In a proper binary tree, the number of external nodes is 1 more than the number of internal nodes, i.e. e = i + 1.

i

e e

Proof Clearly true for BT with one internal node.

i

i

e

i

eee

Mathematical Induction Assume we want to verify the correctness of a statement (P(n)).(1) First, prove that P(n) holds for n=1 (2, 3);(2) assume it holds for an arbitrary n and try to prove it holds for (n+1).

i

i

e

e

e


Clearly true for BT with two internal nodes.

Clearly true for BT with three internal nodes.

Assume true for trees with up to n nodes.

7E = I + 1

should prove Enew = (I+1) + 1 = Inew + 1

In the new tree: Inew = I+1Enew = (E-1)+2 = E+1 = (I+1)+1 = Inew+ 1

8

Assume true for a tree with n nodes: E = I + 1.Now, we want to extend the tree by adding new nodes:

1. Cannot add only one node.• add one internal – impossible! • add one external – that would violate the propertyof proper binary tree

2. Add two externals. In this case, one old externalbecomes internal, so we have:

Inew = I+1Enew = (E-1)+2 = E+1 = (I+1)+1 = I+2

Hence, Enew = Inew + 1

ie i

ee

ii i

eeee

What is the next bigger proper binary tree??


tree with n nodes

next larger proper tree

9

Property 3 The number of external nodes (e) satisfies: (h+1) ≤ e ≤ 2h.

Proof

Property 4 The number of internal nodes (i) satisfies: h ≤ i ≤ 2h-1.Proof

Based on Property 3: h+1 ≤ e ≤ 2h

Based on Property 2: h+1 ≤ i+1 ≤ 2h

h ≤ i ≤ 2h-1

At every level (except last)there is only one external node.

Full binary tree.

If the tree-height (h) is known, what is the number of external nodes?!

If the tree-height (h) is known, what is the number of internal nodes?!


i+1

h=3

10

Property 5 The total number of nodes (n) satisfies: 2h+1 ≤ n ≤ 2h+1-1.

ProofBased on Property 3: (h+1) ≤ e ≤ 2h

Based on Property 4: h ≤ i ≤ 2h-1Based on n=i+e: 2h+1 ≤ n ≤ 2h+1-1

2h+1 ≤ n ⇒ h ≤ (n-1)/2

Property 6 The height (h) satisfies: log2(n+1)-1 ≤ h ≤ (n-1)/2.

Proof Based on Property 5, the following two inequalities hold:

n ≤ 2h+1-1 ⇒ n+1 ≤ 2h+1 ⇒ log2(n+1) ≤ h + 1 ⇒ log2(n+1) - 1 ≤ h


If the number of nodes in a tree is known, what is the height of the tree?!

If the tree-height (h) is known, what is the overall number of nodes?!

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Property 7 The height (h) satisfies: log2(e) ≤ h ≤ e-1.

ProofBased on Property 3: (h+1) ≤ e ≤ 2h

From 1st inequality: h ≤ e-1From 2nd inequality: log2(e) ≤ h

Property 8 The height (h) satisfies: log2(i+1) ≤ h ≤ i.

ProofBased on Property 2: e = i + 1Apply Property 2 to Property 7.


If the number of external nodes is known, what is the height of the tree?!

If the number of internal nodes is known, what is the height of the tree?!

12

Summary of Properties

n = e + i

e = i + 1

(h+1) ≤ e ≤ 2h

h ≤ i ≤ 2h-1

2h+1 ≤ n ≤ 2h+1-1

log2(n+1)-1 ≤ h ≤ (n-1)/2

log2(e) ≤ h ≤ e-1

log2(i+1) ≤ h ≤ i

Number of external, internal, and overall nodes as a

function of tree’s height

Tree’s height as a function of number of external, internal,

of overall nodes

All other expressionscan be obtained from

these 3.


13Binary Tree ADT: Interface

Additional Methods public Position<E> left(Position<E> v) throws…;/* return left child of a node; null if no child exists */

public Position<E> right(Position<E> v) throws…;/* return right child of a node, null if no child exists */

public Position<E> sibling(Position<E> p) throws…;/* returns position of p’s sibling, null if no sibling exists */

Binary Tree ADT – extends Tree ADT, i.e. inherits all its methods:size, isEmpty, iterator, positions, replace, root, parent,children, isInternal, isExternal, isRoot

14Binary Tree: Array-Based Implementation

A

B C

D E F G

Indexing Scheme – for every node v of T, let its index/rank rank(v) bedefined as follows:

• if v is the root: rank(v) = 1• if v is the left child of node u: rank(v) = 2*rank(u)• if v is the right child of node u: rank(v) = 2*rank(u) + 1

A

2 3

4 5 6 7

B C D E F G

1

Advantages – simple implementation, easy access

No element at rank 0! 0 1 2 3 4 5 6 7

How could the parent-rank be

found?!

15

Space Complexity – let use the following notation• n – number of nodes in T• rankmax – maximum value of rank(v)• N – array size (i.e. space usage): N=rankmax+1

1) Best Case: full, balanced tree ⇒ all array slots occupied

rankmax = n ⇒ N = O(n)

2) Worst Case: highly unbalanced tree ⇒ many slots empty

height: h = (n-1)/2

rankmax: rankmax = 2h+1 – 1 = 2(n+1)/2 – 1

required: N = rankmax+1 = 2(n+1)/2 = O(2n)array size

1

2 3

6 7

14 15

rankmax -as if the treeof this height

was full

Binary Tree: Array-Based Implementation

16Array-Based Binary Tree: Performance

Method

position, iteratorsize, isEmpty

replaceroot, parent, children, left, right

hasLeft, hasRightisInternal, isExternal, isRoot

Time

O(n)O(1)O(1)O(1)O(1)O(1)

Run Times – Good! all methods, except iterator and positions run inO(n) time

Space Usage – Poor! (in general)best case (full balanced tree): O(n)worst case (highly unbalanced tree): O(2n)

Explain!

Binary Trees (1) - York · PDF file8 Assume true for a tree with n nodes: E = I + 1. Now, we want to extend the tree by adding new nodes: 1. Cannot add only one node. •add one internal

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