1. Trees are Everywhere family genealogy organizational charts
corporate
government
military
folders/files on a computer
compilers: parse tree
a = (b + c) * d; = a * + d b c
Oct 19, 2001
CSE 373, Autumn 2001

2. First child/next sibling
struct TreeNode {
Object element;
TreeNode *firstChild;
TreeNode *nextSibling; }
C:\
MyMail
cse373
D101
school
pers
hw1
hw2
proj1
coll.h
coll.cpp
poly.h
Oct 19, 2001
CSE 373, Autumn 2001

3. Tree traversal preorder/postorder traversal
TreeNode::printTree(TreeNode tnode) {
tnode.print();
for each child c of tnode
c.printTree(); }
preorder/postorder traversal
int TreeNode::numNodes(TreeNode tnode)
if (tnode == NULL)
return 0;
else
sum = 0;
sum += numNodes(c);
return 1 + sum;
Oct 19, 2001
CSE 373, Autumn 2001

4. Binary trees
A binary tree is a tree where all nodes have at most two children.
struct BinaryNode {
Object element;
BinaryNode *left;
BinaryNode *right; } 1 2 1 3 2 3 4 4 5 6 7 5 6 7
Oct 19, 2001
CSE 373, Autumn 2001

5. Binary Search Trees
Associated with each node is a key value that can be compared.
Binary search tree property:
every node in the left subtree has key whose value is less than the value of the root’s key value, and
every node in the right subtree has key whose value is greater than the value of the root’s key value.
Oct 19, 2001
CSE 373, Autumn 2001

6. Example 5 4 8 1 7 11 3 BINARY SEARCH TREE Oct 19, 2001
CSE 373, Autumn 2001

7. Counterexample 8 5 11 2 7 6 10 18 4 15 20 NOT A BINARY SEARCH TREE 21
Oct 19, 2001
CSE 373, Autumn 2001

8. find
Basic idea: compare the value to be found to the key of the root of the tree.
If they are equal, we are done.
If they are not equal, recurse depending on which half of the tree the value to be found should be in if it is there.
BST::find(const int x, BNode<int> *t) {
if (t == NULL)
return NULL;
else if (x < t->element)
return find(x, t->left);
else if (x > t->element)
return find(x, t->right);
else
return t; // match }
Oct 19, 2001
CSE 373, Autumn 2001