1. Tree
Lecturers : Boontee Kruatrachue Room no Kritawan Siriboon Room no. 913
Text :
Data Structures & Algorithm Analysis in C, C++,…
Mark Allen Weiss, Addison Wesley

2. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

3. Height of node = longest path length from node to leaf
Leaf Height = 0
Empty Tree Height = -1
Tree Height = Root Heigth r 5 A 4 B C 3 D 3 E F G H 2 2 I J K L 1 1 M N
O = Longest Leaf from C O P
P = Longest Leaf from A

4. Why Balanced Tree ?
Tree Efficiency (searching, inserting, deleting,...) → O(Height)
Binary Search Tree
sometimes turn degenerated 6 3 1 4 2 7 5
Linear Linked list or near
Linear Search O(n) →
Binary Search Tree
→ (Perfect) Balanced Tree O(log2n)
Every leaf is at the same level (very rear case)
Balanced Binary Search Tree
Non-Height Balanced / \ - //
Height Balanced 5 1 8 7 2 4 3
→ Hight Balanced Tree O(log2n) → AVL log2n
Every node’s subtrees’ heights differ <= 1

5. Every node’s balance factor = 0
Height Balanced Tree : Every node’s subtrees’ heights differ <= 1
(Nearly Balanced) Balance factor of every node is 0 or 1 or -1
Balance factor of a node : different of left & right subtrees’ hights = Height(Left Subtree) – Height(Right Subtree) - /
Height Balanced -
Height Balanced
Every node’s balance factor = 0 - \ /
Height Balanced 1
0-(-1) 1
0-(-1) -1
-1-0 - / \ //
Non - Height Balanced
2 1-(-1) -1
1-2
1 0-(-1)
-1-0

6. AVL (Height Balanced) Tree
AVLTree Self- rebalancing binary search tree to keep height balance.
Maximum height <= 1.44 log2n : near the ideal complete binary tree.
Search, insertion, deletion ( both average & worst cases) O(log2n)
Insertions & deletions may require 1 or > 3 rotations.
Worst case not much slower than random binary search tree.
Adelson-Velskii
Landis
(1962 Soviet inventors) 5 1 8 7 2 4 3 \ / -
Height Balanced
Smallest AVL tree of height 9
Bad Binary Tree
Balance at root
is not enough. - /
/// // \
\\\ \\
Non-Height Balanced / \ - //

7. Searching in AVL = Searching in BST
AVL is a binary search tree.
Review Searching a BST (Binary Search Tree).
Compare search key against key in node start from root.
Follow associated link (recursively).
Left link if search key < key in node
Right link if search key ≥ key in node
Ex. Search for 3
3 < 5 3 1 4 7 9 2 8 5
3 ≥ 2
3 < 4
found 3

8. Review Inserting in BST
Search to bottom for key. / - 1 7 9 2 8 5
Ex. Insert 3 Ex. Insert 6 1 7 9 2 8 5 / - / - 1 7 9 2 8 5
is NULL
Not found 3
is NULL
Not found 6

9. Review Inserting in BST
Search to bottom for key. / - 1 7 9 2 8 5
NULL
Ex. Insert 3 Ex. ert 6 / - 1 7 9 2 8 5 / - 1 7 9 2 8 5
Not found 3
Not found 6

10. Review Inserting in BST
Search to bottom for key.
Insert node as a new leaf node. / - 1 7 9 2 8 5
NULL
Ex. Insert 3 Ex. Insert 6 3 / - 1 7 9 2 8 5 / - 1 7 9 2 8 5 6
Not found 3
Insert 3 as a new leaf
Not found 6
Insert 6 as a new laef -

11. Inserting in AVL without Rebalancing1 7 9 2 8 5 / -
h = 1
h =-1
h = 2
h = 0
Inserting at short side or equal side → AVL ‘s still height balanced
Insert BST new leaf node.
Inserting node will change the height of some nodes (increase by 1)
Change balance factors along the path to the root. 3
Short Side
Equal Side 6
Long Side
on the path from the newly inserted node to the root only.
Ex. Insert 3 at short side.
Ex. Insert 6 at equal side.
insert 3 at short side
still height balance - 1 7 9 2 8 5 1 7 9 2 8 5 / -
insert 6 at equal side
still height balance - - \ / 3 6

12. Insertion Re-balancing in AVL
h = 4 10 30 45 55 20 50 65 75 87 40 70 85 95 90 60 80
h = 1
Short Side / \ -
h = 3
h = 0
h = 2 x
1st unbalanced x
Insert 15
h = 1 10 30 45 55 20 50 65 75 87 40 70 85 95 90 60 80
h = 2
Short Side / \ -
h = 4
h = 0
h = 3
h = 5 // / \ -
Hight diff
= 2
Long Side
Long Side 15 5 15 25 35 43 46 52 57
Height Unbalanced → no longer AVL !
Height balanced again→ AVL 20 40 60 80
h = 2 - / 90 87 85 95 \ 65 75 70
h = 1 45 55 50 10 30
h = 0 15 x y z
Re-balance
Inserting AVL in a long side.
Causes some node(s) unbalanced (height diff = 2) only up the inserted path.
x = 1st unbalanced node
Need rebalancing the (blue) shaded subtree.
Rebalancing does not change :
height of the subtree
→ ie. Nor height of root.
→ Only some part of the shaded subtree’s balance facters need changing.
h = 4
h = 3

13. 4 Insertion Re-balancing in AVL10 30 20 65 75 87 40 70 85 95 90 60 80 5 15 25 35 - \ / 45 55 50 43 46 52 57 x
Short Side
Long Side 65 55 60 25 15 7 50 20 5 1 3 30 10 70 62 59 51 - \ 45 35 40 46 41 37 32 x
Short Side
Long Side
1. Left_Left
2. Right_Left
4. Left_Right
3. Right_Right
Inserting AVL in a long side. x = 1st unbalanced node , inserted node is at
1. left subtree of left son of x 2. right subtree of left son of x
3. right subtree of right son of x 4. left subtree of right son of x z y x z y x
left son of x
left son of x z y x z y x
right son of x
right son of x

14. Rebalancing Left Subtree of Left Son of x (1st Unbalanced Node)
1st unbalanced x
son of y : z
son of x : y T0 T2 T3 10 30 45 55 20 50 65 75 87 40 70 85 95 90 60 80
h = 2
Short Side // / \ -
h = 4
h = 1
h = 0
h = 3
h = 5 T1
h = 4 10 30 45 55 20 50 65 75 87 40 70 85 95 90 60 80 5 15 25 35
h = 1
Short Side / \ -
h = 3
h = 0 43 46 52 57
Long Side
h = 2 x
Height Unbalanced → no longer AVL !
hight diff
= 2
Insert 15
Long Side 15
x = 1st unbalanced node. Along the path, let y = son of x, z = son of y.
Left subtree of left son of x
Right rotate at x
x down to be y’s (right) son.
y up to replace x’ s place : be son of x’s father.
Re-arrange T0,T1,T2,T3 to preserve the properties of BST.
T0 ≤ z ≤ T1 ≤ y ≤ T2 ≤ x ≤ T3
Fixing some part of the shaded subtree’s balance facters.
Height of the blue shaded subtree doesn’t change -> also the whole AVL‘s.
h = 3
h = 4 80 / 90 87 85 95 \ -
Re-balance 20 40 60 x y z
h = 2 T0 10 30
h = 1 - T1 15
h = 0 - \ / T2 45 55 50 -
h = 1 T3 65 75 70 -
h = 1
Height balanced again→ AVL

15. Proof : Rebalancing Area’ Height’s unchange (Left Subtree of Left son of x)
Insert at a long side T0
x = 1st unbalanced node, y = son of x, z = son of y
height(T1) = ? When height(T0) = h.
= h-1 or h ∵ heightDiff (T0,T1 ) ≤1 & ≠ h+1 ∵ T1 : short side, T0 : long side
= h ∵ After inserting , z : balanced ∴ heightDiff ≤1
height(T2) = h+1 ∵ After inserting , y : balanced
height(T3) = h+1 ∵ After inserting , x : unbalanced z y x T1 T2 T3 T0
h+3
h+2 ?
h+1 h
Rebalancing does not change heights of the subtree → ie. Nor of root.
So only fix balance facters of inserted path to y and of x.
Long Side
h+3 h
h+1
h+2
h-1/h
h+1/h+2 - /
Insert at T0 z y x T1 T2 T3 T0
h+4
h/h+1 //
1st unbalanced
1st Diff > 1
balanced
H.Diff ≤1
Right Rotate
at x
Long Side

16. Proof : Rebalancing Area’ Height’s unchange (Left Subtree of Left son of x)2
Insert at a long side T1
x = 1st unbalanced node, y = son of x, z = son of y
height(T0) = ? When height(T1) = h. = h-1 or h ∵ heightDiff (T0,T1 ) ≤1 & ≠ h+1 ∵ T0 : short side, T1 : long side = h ∵ After inserting , z : balanced ∴ heightDiff ≤1
height(T2) = h+1 ∵ After inserting , y : balanced
height(T3) = h+1 ∵ After inserting , x : unbalanced h
h+1
h+2
h+3 ? z y x T1 T2 T3 T0
Rebalancing does not change heights of the subtree → ie. Nor of root.
So only fix balance facters of inserted path to y and of x.
Long Side z y x T1 T2 T3 T0
h+1
h+2
h+3
h+4 h
1st unbalanced
balanced \ / // -
1st Diff > 1
H.Diff ≤1
Right Rotate
at x
Insert at T1
h-1/h
h/h+1
h+1/h+2
Long Side

17. 4 Kinds of Rebalancing in AVL
x = 1st unbalanced node, y = son of x, z = son of y
Left subtree of left son of x
2. Right subtree of left son of x 2 z y x T1 T2 T3 T0 or z y T1 T0 x T3 T2 or z y x T2 T0 T3 T1 or y z T1 T0 x T3 T2 or 1
3. Right subtree of right son of x
4. Left subtree of right son of x 2 z y x T2 T1 T0 T3 or z y T2 T3 x T0 T1 or z y x T1 T3 T0 T2 or y z T2 T3 x T0 T1 or 1

18. Proof : Rebalancing Area’ Height’s unchange (Right Subtree of Left son of x)
Insert at a long side T1
x = 1st unbalanced node, y = son of x, z = son of y
height(T2) = ? When height(T1) = h. = h-1 or h ∵ heightDiff (T2,T1) ≤1 & ≠ h+1 ∵ T2 : short side, T1 : long side = h ∵ After inserting , z : balanced ∴ heightDiff ≤1
height(T0) = h+1 ∵ After inserting , y : balanced
height(T3) = h+1 ∵ After inserting , x : unbalanced
Rebalancing does not change heights of the subtree → ie. Nor of root.
So only fix balance facters of inserted path to z and of y.
h+3 -
h+2 \
Insert
at T1
Left Rotate
at y h
h+1
h+4 z y x T2 T0 T3 T1 // / 1 2
h-1/h
h/h+1
h+1/h+2
Right Rotate
atx
Long Side
1st unbalanced
1st Diff > 1
balanced
H.Diff ≤1

19. Proof : Rebalancing Area’ Height’s unchange (Left Subtree of Right Son of x )
Insert at a long side T1
x = 1st unbalanced node, y = son of x, z = son of y
height(T2) = ? When height(T1) = h. = h-1 or h ∵ heightDiff (T1,T2) ≤1 & ≠ h+1 ∵ T2 : short side, T1 : long side = h ∵ After inserting , z : balanced ∴ heightDiff ≤1
height(T3) = h+1 ∵ After inserting , y : balanced
height(T0) = h+1 ∵ After inserting , x : unbalanced
Rebalancing does not change heights of the subtree → ie. Nor of root.
So only fix balance facters of inserted path to z and of y.
Insert
at T1
Right Rotate
at y h T2
h+1 x T0 \\ 2 y T3 - z T1
h+3 \
h+2
Left Rotate
at x
h-1/h
h/h+1
h+1/h+2
Long Side
h+4 / 1
1st unbalanced
1st Diff > 1
balanced
H.Diff ≤1

20. Single Right Rotation & Single Left Rotation Algorithms
1. Left subtree of left son of x
// Single Right Rotation
LeftLeftRotate(node *px)
node* py = px->left
px->left = py->right
py->right = px
return py z y T1 T0 x T3 T2 or z y x T1 T2 T3 T0 or px py
3. Right subtree of right son of x
// Single Left Rotation
RightRightRotate(node *px)
node* py = px->right
px->right= py->left
py->left = px
return py z y T2 T3 x T0 T1 or z y x T2 T1 T0 T3 or px py

21. Single Right Rotation & Single Left Rotation Algorithms
// Double Rotation :
// Case 2 : Left Rotation then Right Rotation
RightLeftRotate(node *px)
node* py = px->left
px->left = RightRightRotate(px->left)
py = LeftLeftRotate(px)
return py
2. Right subtree of left son of x z y x T2 T0 T3 T1 or 1 2 px py
4. Left subtree of right son of x
// Double Rotation :
// Case 4 : Right Rotation then Left Rotation
LeftRightRotate (node *px)
node *py = px->right
px->right = LeftLeftRotate(px->right)
py = RightRightRotate(px)
return py z y x T1 T3 T0 T2 or 1 2 px py

22. B ตัวค่ากลาง เป็น root ใหม่ เสมอ
Single Rotation in AVL
x = 1st unbalanced node, y = son of x, z = son of y A < B < C T0<T1<T2<T3
Left subtree of left son of x A B C A B C T1 T0 z y x T2 T3 or
h+3 - / h
h+1
h+2 z y T1 T0 x T3 T2
h+3 - \ z y T1 T0 x T3 T2
h+1 h
h+2
สรุปรูปผลลัพธ์
B ตัวค่ากลาง เป็น root ใหม่ เสมอ Tj Tj
Height(Tj) = h ตัวเดียว Tj คือตัวที่คู่กับ Ti ที่ insert node ใหม่เข้าไป
นอกนั้น height(T) = h+1 หมด
2. Right subtree of right son of x z y x T2 T1 T0 T3 or A B C A B C
h+3 - \ h
h+1
h+2 z y T2 T3 x T0 T1
h+3 - \ h
h+1
h+2 z y T2 T3 x T0 T1 Tj Tj

23. Double Rotations in AVL
x = 1st unbalanced node, y = son of x, z = son of y A < B < C T0<T1<T2<T3
Right subtree of left son of x z y x T2 T0 T3 T1 1 2 or A B C A B C
h+3 -
h+2 \ y z T1 T0 x T3 T2
h+1 h
h+3 / -
h+2 y z T1 T0 x T3 T2 h
h+1
สรุปรูปผลลัพธ์
B ตัวค่ากลาง เป็น root ใหม่ เสมอ Tj Tj
Height(Tj) = h ตัวเดียว Tj คือตัวที่คู่กับ Ti ที่ insert node ใหม่เข้าไป
นอกนั้น height(T) = h+1 หมด
3. Left subtree of right son of x T3 z y x T1 T0 T2 or 1 2 A B C A B C
h+3 \ -
h+2 y z T2 T3 x T0 T1 h
h+1
h+3 -
h+2 / y z T2 T3 x T0 T1
h+1 h Tj Tj

24. What is the Result AVL after insert 70 ?50 30 10 60 - 35 45 40 \ 1 5 3 7 15 25 20 55 65 70

25. Result AVL after Insert 7050 30 10 60 - 35 45 40 1 5 3 \ 7 15 25 20 55 65 70 \\ \\ 50 30 10 T3 55 65 60 70 \ - T2 35 45 40 1 5 3 7 T1 15 25 20

26. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

27. Which Representations ?
typedef int T;
struct node{
T data;
struct node *left, *right; };
typedef struct node node;
Dynamic 9 3 5 6 1 r
Linked Array
Sequential (Implicit) Array G A C L M F I D B 13 14 6 2 11 12 5 9 10 4 7 8 3 1
left
data
right
father 1 9 2 3 5 4 6 -1 2 -1 4 1 3 4
root 4 2 -1 2 -1 A B C D F G I L M 1 2 3 4 5 6 7 8 9 10 11 12 13 14

28. Sequential Array (Implicit Array)G A C L M F I D B 13 14 6 2 11 12 5 9 10 4 7 8 3 1 A B C D F G I L M 1 2 3 4 5 6 7 8 9 10 11 12 13 14
ง่าย
ต้องกะขนาด array
เพิ่มขึ้นอีก 1 level # ที่เพิ่มมากกว่าที่มีอยู่เดิมอยู่ (คิดเต็มที่ full binary tree)
ถ้ากะขนาด array ถูก และ tree ใช้พื้นที่ส่วนใหญ่ของ array เช่น almost complete binary tree) จะ save space เพราะไม่ต้องมีฟิลด์ left, right, father

29. Linked Array 1 9 2 3 5 4 6 r -1 2 -1 4 1 3 4 root 4 2 -1 2 -1 6 3 9 1
left
data
right
father 1 9 2 3 5 4 6 9 3 5 6 1 r -1 2 -1 4 1 3 4
root 4 2 -1 2 -1
ต้องใช้ left, right (or father) ฟิลด์
flexible กว่า ดึง node มาใช้จาก pool กลาง available list
จำนวน node ถูก limited โดยขนาดของ array
ตำแหน่งของ node ไม่เจาะจงเหมือน sequential

30. Dynamic 9 3 5 6 1 r
จำนวน node ถูกจำกัดโดย memory

31. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

32. n-ary Trees
N-ary Tree : Each node has at most n children. A B C D E F G
Ternary Tree.

33. Generic Tree
Generic Tree : Each node has unrestricted number of children.
root T3
T10 T4 T1 T2 B C D A E F G H I J K L M N P Q

34. Implementation of Generic TreeB C D E F G
typedef int T;
struct treeNode{
T data;
struct treeNode *firstChild
struct treeNode *nextSibling; };
typedef struct treeNode treeNode; H I J K L M N P Q B C D A E F G H I J K L M N P Q
firstChild
nextSibling
nextSibling
nextSibling
nextSibling
nextSibling
firstChild
firstChild
firstChild
firstChild
firstChild
nextSibling
nextSibling

35. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

36. Multiway Search Tree
Multiway Tree order n : has at most n subtrees (has 0, 1,... or n subtrees) (#max. sons = n)
Order 4: มีลูกได้มากที่สุด 4 ตัว
Root has maximum sons = 4
How many keys of the father of 4 sons ?
Multiway Search Tree : for every key K
right decendents > K
left decendents ≤ K 3
So, for order 4, #maximum key = 3
For order n :
#max. sons = n
#max. Keys = n-1
leaf : node ที่ไม่มีลูกเลย
Semi-leaf node : มีลูกอย่างน้อย 1 ตัวเป็น NULL
Semi-leaf
Full node : node ที่มีจำนวน key เต็มที่
Full node
Order 4
#max. sons = 4
#max. keys = 3
leaf
= NULL

37. Multiway Search Tree
Multiway Tree order n : has at most n subtrees (has 0, 1,... or n subtrees) (#max. sons = n)
Order 4: มีลูกได้มากที่สุด 4 ตัว
Root has maximum sons = 4
How many keys of the father of 4 sons ?
Multiway Search Tree : for every key K
right decendents > K
left decendents ≤ K 3
So, for order 4, #maximum key = 3
For order n :
#max. sons = n
#max. Keys = n-1
leaf : node ที่ไม่มีลูกเลย
Semi-leaf node : มีลูกอย่างน้อย 1 ตัวเป็น NULL
Semi-leaf
Full node : node ที่มีจำนวน key เต็มที่
Full node
Order 4
#max. sons = 4
#max. keys = 3
leaf
= NULL

38. Multiway Search Tree

39. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

40. Top Down Tree Top Down Tree : Non full node must be leaf.
Order n : has at most n subtrees (has 0, 1,... or n subtrees) (#max. sons = n)
Fill up the existing nodes before creating a new node !
→ have fewer nodes as possible
→ fewer nodes : shorter height !
Order 3
#max. sons = 3
#max. keys = 2
Insert 57
Must be with 55
Not in a new node. 57

41. Unbalanced Tree

42. Balanced Tree
Balanced Tree : every leaf node is at the same level (Perfect balance , perfectly height-balanced)

43. Balanced Tree
Balanced Tree : every leaf node is at the same level (Perfect balance , perfectly height-balanced)

44. Tree 2 AVL Tree Height Balanced Tree Which Representions ? n-ary Tree
Generic Tree
Multiway Search Tree
Top Down Tree
B-Tree
Tree Definitions
Binary Tree
Traversals
Binary Search Tree
Representations
Application : Expression Tree

45. Balance Trees
Balanced Tree : every leaf node is at the same level (Perfect balance , perfectly height-balanced)
Order 3
#max. sons = 3
#max. keys = 2

46. Multi-way search tree order n #max son = n, #max keys = n-1
B-Trees
B-tree order n
Multi-way search tree order n #max son = n, #max keys = n-1
Balanced tree
ทุก node ยกเว้น root node มีจำนวน non-empty subtree #min sons = n/2 (Half Full)
EX.
order 2 : #max sons =2 , #min sons = 2/2 = 1
∴ Binary ที่เป็น complete binary tree จะเป็น B-tree
order 5 : #max sons = 5 , #min sons = 5/2 = 2.5 = 3
order 201 : #max sons = 201 , #min sons = 201/2 = 101

47. order 5 #max son = 5, #max keys = 4
B-Tree Variation
order 5 #max son = 5, #max keys = 4
Half Full (ยกเว้น root node) #min sons = 5/2 = 2.5 = 3 B-tree -> variations
internal nodes : เก็บเฉพาะ key เพื่อประหยัดพื้นที่, order กำหนดจำนวนลูก
external nodes : เก็บ data ทั้ง record, ค่า L กำหนดจำนวน record/node
ถ้า L = 5 : #max data = 5, #min data = 5/2 = 2.5 = 3

48. B-Tree (Variation) Insertion
Insert at leaf:
until full, if full -> split
B-Tree (Variation) Insertion
Insertion consider #maximum
order 5 : #max son = 5,
Internal : #max keys = 4,
#min sons = 5/2 = 3
Leaf L = 5: #max data = 5,
#min data = 5/2 = 3

49. B-Tree (Variation) Insertion
Insert at leaf:
until full, if full -> split
B-Tree (Variation) Insertion
order 5, #max key = 5-1=4
L=5 #leaf max data = 5
Insertion consider #maximum
order 5 : #max son = 5,
Internal : #max keys = 4,
#min sons = 5/2 = 3
Insertion consider #maximum
Leaf L = 5: #max data = 5,
#min data = 5/2 = 3

50. B-Tree (Variation) Insertion
Insert at leaf:
until full, if full -> split
B-Tree (Variation) Insertion
order 5, #max key = 5-1=4
L=5 #leaf max data = 5
Insertion consider #maximum
order 5 : #max son = 5,
Internal : #max keys = 4,
#min sons = 5/2 = 3
Leaf L = 5: #max data = 5,
#min data = 5/2 = 3

51. B-Tree (Variation) Insertion
Insert at leaf:
until full, if full -> split
B-Tree (Variation) Insertion
order 5, #max key = 5-1=4
L=5 #leaf max data = 5
Insertion consider #maximum
order 5 : #max son = 5,
Internal : #max keys = 4,
#min sons = 5/2 = 3
Leaf L = 5: #max data = 5,
#min data = 5/2 = 3

52. B-Tree (Variation) Deletion
Delete at leaf:
until low, if low -> borrow
B-Tree (Variation) Deletion
Deletion consider #minimum
Internal order 5 : #min sons = 5/2 = #min keys = 2
Leaf L = 5: #max data = 5, #min data = 5/2 = 3
Borrowing
1. ยืมพี่น้อง(ผ่านพ่อ) เช่น del 49
2. ยืมพ่อ ต้อง consolidate เช่น del 84
พ่อไม่มี พ่อยืม
-พี่น้องพ่อ(ผ่านปู่)
. . .
Delete at leaf:
until low, if low -> borrow

53. B-Tree (Variation) Deletion
Delete at leaf:
until low, if low -> borrow
B-Tree (Variation) Deletion
Deletion consider #minimum
Internal order 5 : #min sons = 5/2 = #min keys = 2
Leaf L = 5: #max data = 5, #min data = 5/2 = 3
Borrowing
1. ยืมพี่น้อง(ผ่านพ่อ) เช่น del 49
2. ยืมพ่อ ต้อง consolidate เช่น del 84
พ่อไม่มี พ่อยืม
-พี่น้องพ่อ(ผ่านปู่)
. . .

54. B-Tree (Variation) Deletion
Delete at leaf:
until low, if low -> borrow
B-Tree (Variation) Deletion
Deletion consider #minimum
Internal order 5 : #min sons = 5/2 = 3 #min keys = 2
Leaf L = 5: #max data = 5, #min data = 5/2 = 3
Borrowing
1. ยืมพี่น้อง(ผ่านพ่อ) เช่น del 49
2. ยืมพ่อ ต้อง consolidate เช่น del 84
พ่อไม่มี พ่อยืม
-พี่น้องพ่อ(ผ่านปู่)
. . .