Michael Tsai 2017/04/11
Reference
• Fundamentals of Data Stru ctures in C, 2nd Edition, 2008
• Chapter 5
• Horowitz, Sahni, and Ande rson-Freed
Reference
• Data Structures and Algor ithms Made Easy, Second E dition, 2011, CareerMonk Publications, by Karumanc hi
• Chapter 6.11
樹
http://www.ahneninfo.com/de/ahnentafel.htm
In the CS world, we usually draw the tree upside-down.
Root
Leaves What are the
properties of a tree?
A nonlinear, hierarchical way to represent data.
Definition
• Definition: A tree is a finite set of one or more no des such that
• (1) There is a specially designated node called the root.
• (2) The remaining nodes are partitioned into disjoi nt sets , where each of these sets is a tree.
• (3) are called the subtrees of the root.
• Note that the above is a recursive definition.
• A node with no subtree, is it a tree?
• Is “No node” (null) a tree?
• Is the “graph” on the right a tree?
•
Tree Dictionary
• Root
• Node/Edge (branch)
• Degree (of a node):
The number of subtrees of a node
• Leaf/Terminal node:
its degree=0
• Parent/Children
• Siblings
they have the same parent.
• Ancestors/Descendants
A
B C D
E F G H I J
K L M
Tree Dictionary
• Level/depth (of a node):
The number of branch to reach tha t node from the root node.
(i.e., root is at level 0)
• Height (of a tree):
The number of levels in a tree
(Note that some definitions start from level 1)
• Size (of a tree):
The number of nodes in a tree
• Weight (of a tree):
The number of leaves in a tree
• Degree (of a tree):
The maximum degree of any node in a t ree
A
B C D
E F G H I J
K L M
Level 0
Level 1
Level 2 Level 3
Representing a tree with array
• What to store?
• The data associated with each node
• Array method
• Store the data sequentially according to the level of the node
• In the array:
• How to find the parent of a node?
• How to find the children of a node?
A
B C D
E F G H I J
Max degree of the tree=3
A B C D E F G H I J
0 1 2 3 4 5 6 7 8 9 1
0
1 1
1 2
[1]-[3]
[0]
[4]-[6] [7]-[9] [10]-[12]
Representing a tree with array
• Assume degree (of the tree)=d (The example on the right uses d=3)
• How to find the parent of a nod e?
• Observation:
For node with index i, its pare nt’s index is
• How to find the children of a n ode?
• Observation:
For node with index i, its chil dren’s indices range from ???
•
A
B C D
E F G H I J
[1]-[3]
[0]
[4]-[6] [7]-[9] [10]-[12]
Representing a tree with array
• Downside?
• If there are many nodes with degree less than d,
• then we waste a lot of space in the array.
A
B C D
E G H J
Max degree of the tree=3 K
I
A B C D E G H I J K
0 1 2 3 4 5 6 7 8 9 1 0 1
1 1
2 … 3 1
Waste of space!
Representing a tree with linked structure
• Assume degree =3 struct TreeNode{
char data;
struct TreeNode* child1;
struct TreeNode* child2;
struct TreeNode* child3;
};
A
B C D
H I A
B \0 \0 \0 C \0 \0 \0 D \0
H \0 \0 \0 I \0 \0 \0
\0 == NULL
Not actually solving the prob?
Wasting the space to store
pointers instead
Representing a tree with linked structure
A
B C … ㄅ
1 2 d
…
• Assume degree of tree = d, size o f the tree = n
• How many pointers are null?
• Total number of pointers:
• Number of branch? n-1.
•
Data child 1 child 2 child 3 … child k
Number of wasted
pointers Better if
smaller!
左小孩 - 右兄弟姊妹 表示法
• Left child-right sibling representation
• Inspired by observations:
• 1. Each node has a leftmost child ( 是廢話 )
• 2. Each node has only a immediately-right sibling ( 也是廢話 )
Data left child Right sibling
Converting to LCRS tree
A
B C D
E F G H I J
K L M
A
B C D
E F G H I J
K L M
LCRS tree
• It’s always a degree-two tree
!
• Converting to from an arbitrar y-degree tree to a degree-two tree!
• Root does not have a right chi ld (because root in the origin al tree does not have a siblin g)
A
B C D
E F G H I J
K L M
A B
C
D E
F
G
H I M J
K
L
Binary Tree
• Definition: 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 t he right subtree.
• According to this definition:
• Note: “null” (no node) is a valid tree.
• Note: the order of the children (left or right) is meaningful.
一些證明
• 證明 : 用歸納法
• i=0 時 , 為 root 那一層 , 所以只有一個 node, 也就是最 多有個 node. ( 成立 )
• 假設 i=k-1 的時候成立 level k-1 最多有個 node
• 那麼 i=k 的時候 , 最多有幾個 node?
• 因為是 binary tree, 所以每個 node 最多有兩個 children
• 因此最多有 node ( 得證 )
•
在 level i 的 node 數目最多為 (root 在 level 0)
兩些證明 ( 誤 )
• 證明 :
• 利用前一頁的結果
• 則總共 node 數目最多為
• . 喔耶 .
•
一棵 height 為 k 的 binary tree, 最多有個 node, .
三些證明 ( 誤 )
• 證明 :
• 假設 n 為所有 node 數目 , 為 degree 1 的 node 數目 ,
• 則 . (1)
• 假設 B 為 branch 的數目 , 則 . (2)
• 而且 (3). ( 只有 root 沒有往上連到 parent 的 branch, 其他的 node 正好每個人一個 )
• (2) 代入 (3) 得 (4)
• (4) 減 (1) 得 . 喔耶 .
•
對於任何不是空的 binary tree, 假設為 leaf node 數目 , 為 degree 2 的 node 數目 , 則 .
Full binary tree
• Definition: a full binary tree of depth k is a bina ry tree of depth k having nodes, .
• In other words, the maximum size tree of depth k (f ull)
• All nodes except the leaves have two children
•
1
2 3
4 5 6 7
Full binary tree of depth 3
Complete binary tree
• Definition: 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.
• All leaves at level k-1 and k-2 (the last two levels) have no “missing ones
”
1
2 3
4 5
1
2 3
4 5 6
1 2
1
2 3
5 6
Yes Yes Yes No
1 2
4 5
No
Height of a complete binary tree
• Exercise: if a complete binary tree has n node, wha t’s the height of the tree?
• Hint: a full binary tree of height k has nodes
• Hint: what is the minimum size of such a tree in te rms of k?
• Hint: use a ceiling or floor function
•
Binary Tree Traversal
• How do we traverse each node in the given binary tree?
• When we reach a certain node, there are three possible act ions:
1. Go to left child (use L to represent left branch)
2. Go to right child (use R to represent right branch)
3. Process the data of this node (use V to represent visiti ng this node)
1
2 3
4 5 6 7
Binary Tree Traversal
• Usually L takes place before R.
• Then we have 3 possibilities:
• VLR: preorder
• LVR: inorder
• LRV: postorder
• Preorder: ABCGE
• Please write down the results of inorder & postorde r traversals.
A
B E C G
Recursive traversal
• Use recursive implementation for tree traversal:
void inorder(treePointer ptr) { inorder(ptr->leftChild);
visit(ptr);
inorder(ptr->rightChild);
}
A
B E C G
How about non-recursive? (iterative)
• Use a stack to help:
for(;;) {
for(;node;node=node->leftChild) push(node);
node=pop();
if (!node) break;
printf(“%s”, node->data);
node=node->rightChild;
}
A
B E C G
Arithmetic Expression
• Example: 1+2*3-5/(4+5)/5
• We have:
• Operand – 1, 2, 3, 5, 4, 5, etc.
• Operator – + * - /
• Parenthesis– (,)
• Feature 1: left-to-right associativity
• Feather 2: infix: operator is between two operands
• Association order is according to the priority of the oper ator
• Example: multiplication’s priority is larger than additio n
Alternative expressions
• Postfix: put the operator after the two operands
• Example
• 2+3*4 2 3 4 * +
• a*b+5 ?
• (1+2)*7 ?
• a*b/c ?
• (a/(b-c+d))*(e-a)*c?
• a/b-c+d*?
• e-a*c?
Binary tree with arithmetic expression
• Every arithmetic expression can be converted to an expression tree
• Preorder prefix
• Inorder infix
• Postorder postfix
• Exercise:
• Draw the arithmetic expression tree of (a/(b-c+d))*(e-a)*c
*
- 5
2 3
Level-order traversal
• What if we use a queue to help with the traversal?
add(ptr);
for(;;) {
ptr=delete();
if (ptr) {
printf(“%s”, ptr->data);
if (ptr->leftChild) add(ptr->leftChild);
if (ptr->rightChild) add(ptr->rightChild);
} else break;
}
A
B E C G
• Problem: looking for the grade of a particular stud ent in the university database.
• Assumptions:
• We know the student ID (key)
• Use the key to find the location where the data is stored
• Frequent addition of new students
• Frequent removal of students who dropped out
• Definition: A binary search tree is a binary tree. It may be empty. If it is not empty then it satisfies the following pr operties:
• 1. The root has a key.
• 2. The keys (if any) in the left subtree are smaller than th e key in the root
• 3. The keys (if any) in the right subtree are larger than th e key in the root
• 4. The left and right subtrees are also binary search trees
• (Hidden) All keys are distinct.
• Note that the definition is recursive. root.key
<root.key >root.key
• Are these binary search trees?
• What’s next when we find the node?
20
15 25
12 10 22
5 40
2
60
70
65 80
Yes
No Yes
No. 55
HW1 65
HW2 65
HW3 空
Extra -20000
BST struct definition
struct BinarySearchTreeNode { int data;
struct BinarySearchTreeNode *left;
struct BinarySearchTreeNode *right;
};
struct TreeNode* find(struct TreeNode* root, int data) { if (root==NULL) return NULL;
if (data==root->data) return root;
if (data<root->data) return find(root->left,data);
return find(root->right, data);
}
• Time complexity = O(??)
• A: O(h), h: height of the tree.
• Worst case: Average case:
• 12 22
10 15 25
Binary Search Tree
Algorithm usually is like:
(1) If the key matches the key of this node, then we process and return.
(2) If key is larger or smaller, then use a
recursive call to process left or right branch.
Other operations
• Q: How to find the largest or smallest key in the BST?
• A: Keep going right(right), until reaching NULL (a lea f).
• Q: How to list all keys in a binary search tree in ascen ding order?
• A: Perform inorder traversal of the BST.
20
12 22
10 15 25
• Search if there exists the same key in the BST (Recall the rule: each key in the BST is unique)
• If not found, insert at the last location (where we cannot find the key)
• Insert: 11
20
12 22
10 15 25
1 1
if (root==NULL) {
root=(struct BinarySearchTreeNode*)malloc(sizeof(struct BinarySea rchTreeNode));
if (root==NULL) {
printf("Error\n");
exit(-1);
}
root->data=data;
root->left=NULL;
root->right=NULL;
}else{
if (data<root->data)
root->left=Insert(root->left,data);
else if (data>root->data)
root->right=Insert(root->right,data);
}
return root;
}
If larger or smaller, use a recursive call to
process.
Return to prev. level
Finding
NULL means we have
reached a leaf and the key is not found. insert at this location.
• First find the location of the node
• Then, according to the conditions:
• The node with the key
has no branch (degree=0)
• Remove the node and then done!
20
12 22
10 15 25
1 1
• If the node with the key has one branch (degree=1)
• Then get its only child and attach to the parent
• Example: remove 25
• Q: How to remember the pointer?
• (return to the previous level to process, similar to slide #37)
• (Karumanchi p.152)
20
12 22
10 15 25
1 1
• What if both branches of the node with the key exist (degree=2)?
• Example: remove 12
• Find the largest node of the left branch (or the smallest of
the right branch)
• Remove that node and move it to where the node with the key was.
• Q: What if that node still has child node(s)?
20
12 22
10 15 25
1 1
A: the re wo uld b
e onl y one
child .
printf(“error\n”);
return NULL;
} else if (data < root->data)
root->left=delete(root->left, data);
else if (data > root->data)
root->right=delete(root->right, data);
else { // data == root->data
if (root->left && root->right) { //two children temp=findmax(root->left);
root->data=temp->data;
root->left=delete(root->left,root->data);
}else{ // one child or no child temp=root;
if (root->left==NULL) root=root->right;
if (root->right==NULL) root=root->left;
free(temp);
}
return root; } }
smaller, use a recursive call to
process.
If we have found the key, process it here.
Return to the prev.
level. If we
delete the
current node, we can use this to connect the
parent node to a child
node.
