第6讲(树、二叉树)

Iterative Program
void iter_inorder ( tree_ptr tree ) { Stack S = CreateStack( MAX_SIZE ); tree_ptr p = tree; do { while ( p != NULL ) { Push ( S , p ) ; p = p->Left; } if ( ! isEmpty(S) ) { p = Pop( S ); visit ( p->Element ); p = p->Right; } } while (p != NULL || ! isEmpty(S) ) ; }
CHAPTER 4
TREES
§1 Preliminaries
1. Terminology
Pedigree Tree ( binary tree ) Lineal Tree
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§1 Preliminaries
【Definition】A tree is a collection of nodes. The collection can be empty; otherwise, a tree consists of (1) a distinguished node r, called the root; (2) and zero or more nonempty (sub)trees T1, , Tk, each of whose roots are connected by a directed edge A
B
F
C
G H M
D
I J
L
Exercise: height of ni ::= length of the longest path from ni to a leaf. p.139 4.1-4.3 Height(leaf) = 0, and height(D) = 2.
N N
M
N N
Note: The representation is not unique since the children in a tree can be of any order.
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〖Exercise 1〗
Draw the FirstChild-NextSibling representation of the following tree. A B A F G K K L M N L M
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B
F G
C
H
F I
D
J
E
C H I N J D E
〖Exercise 2〗
Draw the FirstChild-NextSibling representation of the following forest. A A B C D F I E B G J H C D I J F G H E
B
E F
A
C
G
H M
D
I
J
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FirstChild-NextSibling Representation
Element FirstChild NextSibling
A B E F C G H D I J
§1 Preliminaries
A
N
B
C
D
N
E
F
N N
G
N N
H
I
N
J
N N
K
L
M
K
N
L
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path from n1 to nk ::= a (unique) sequence of nodes n1, n2, …, nk such that ni is the parent of ni+1 for 1 i < k. length of path ::= number of edges on the path. depth of ni ::= length of the unique path K from the root to ni. Depth(root) = 0.
Constructing an Expression Tree (from postfix expression)
+
A
§2 Binary Trees

B C
D
〖Example〗 ( a + b ) * ( c * ( d + e ) ) = a b + c d e + * *
+ a T2 T2 a a
〖Exercise 4〗
Give the inorder, preorder and postorder traversal of the following binary tree:
A
B
C
D
E
F
G
inorder : D B E I H A F C G J preorder: A B D E H I C F G J postorder : D I H E B F J G C A
J
H I
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〖Exercise 5〗
Suppose the preorder traversal of a binary tree is abdgcefh, and the inorder traversal is dgbaechf, draw the binary tree and give the postorder traversal of it.
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§1 Preliminaries
degree of a node ::= number of subtrees of the node. For example, degree(A) = 3, degree(F) = 0.
A
B
E K L F
C
G H M
D
I J
max de gree (node) degree of a tree ::= node tree For example, degree of this tree = 3.
〖Example〗 infix expression: A+B CD
+
A
Exercise: p.140 4.8 Given an

B
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D C
Then inorder traversal A + B C D postorder traversal A B C D + preorder traversal + A B C D
Postorder Traversal
void postorder ( tree_ptr tree ) { if ( tree ) { for (each child C of tree ) postorder ( C ); visit ( tree ); } }
Levelorder Traversal
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1 2 3 4
5 6 7
12 3 4 5 6 7
§2 Binary Trees
Inorder Traversal
void inorder ( tree_ptr tree ) { if ( tree ) { inorder ( tree->Left ); visit ( tree->Element ); inorder ( tree->Right ); } }
descendants of a node ::= all the nodes in its subtrees.
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2. Implementation
§1 Preliminaries
List Representation
A
B E K L F C G H M D I J
(A) ( A ( B, C, D ) ) the each ( A ( B ( E, FSo ), C ( Gsize ), D of ( H, I, J )node )) depends on the number of ( A ( B ( E ( K, L ), Fbranches. ), C ( G ), D ( H ( M ), I, J ) ) ) Hmmm... That’s not good. K L
A B C F
*
/
D F
E
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§2 Binary Trees
Tree Traversals —— visit each node exactly once Preorder Traversal
void preorder ( tree_ptr tree ) { if ( tree ) { visit ( tree ); for (each child C of tree ) preorder ( C ); } }
parent ::= a node that has subtrees.
children ::= the roots of the subtrees of a parent. siblings ::= children of the same parent. leaf ( terminal node ) ::= a node with degree 0 (no children).
Note:
Subtrees must not connect together. Therefore every node in the tree is the root of some subtree. There are N 1 edges in a tree with N nodes. Normally the root is drawn at the top.