Lecture 8 tree traversal

Lecture 8
Tree Traversal
Abirami Sivaprasad

Binary Tree Traversal Methods
Preorder
Inorder
Postorder

Preorder Traversal
void preOrder(treePointer ptr)
{
if (ptr != NULL)
{
visit(t);
preOrder(ptr->leftChild);
preOrder(ptr->rightChild);
}
}

Preorder Example (Visit = print)
a
b c
a b c

Preorder Example (Visit = print)
a
b c
d e
f
g h i j
a b d g h e i c f j

Preorder Of Expression Tree
+
a b
-
c d
+
e f
*
/
Gives prefix form of expression!
/ * + a b - c d + e f

Inorder Traversal
void inOrder(treePointer ptr)
{
if (ptr != NULLL)
{
inOrder(ptr->leftChild);
visit(ptr);
inOrder(ptr->rightChild);
}
}

Inorder Example (Visit = print)
a
b c
b a c

Inorder Example (Visit = print)
a
b c
d e
f
g h i j
g d h b e i a f j c

Inorder By Projection (Squishing)
a
b c
d e
f
g h i j
g d h b e i a f j c

Inorder Of Expression Tree
+
a b
-
c d
+
e f
*
/
Gives infix form of expression
ea + b * c d / + f-

Postorder Traversal
void postOrder(treePointer ptr)
{
if (ptr != NULL)
{
postOrder(ptr->leftChild);
postOrder(ptr->rightChild);
visit(t);
}
}

Postorder Example (Visit = print)
a
b c
b c a

Postorder Example (Visit = print)
a
b c
d e
f
g h i j
g h d i e b j f c a

Postorder Of Expression Tree
+
a b
-
c d
+
e f
*
/
Gives postfix form of expression!
a b + c d - * e f + /

Binary Tree Construction
Suppose that the elements in a binary tree
are distinct.
Can you construct the binary tree from
which a given traversal sequence came?
When a traversal sequence has more than
one element, the binary tree is not
uniquely defined.
Therefore, the tree from which the
sequence was obtained cannot be
reconstructed uniquely.

Binary Tree Construction
Can you construct the binary tree,
given two traversal sequences?
Depends on which two sequences are
given.

Inorder And Preorder
• inorder = g d h b e i a f j c
• preorder = a b d g h e i c f j
• Scan the preorder left to right using the
inorder to separate left and right subtrees.
• a is the root of the tree; gdhbei are in the
left subtree; fjc are in the right subtree.
a
gdhbei fjc

• b is the next root; gdh are in the left
subtree; ei are in the right subtree.
a
gdhbei fjc
a
gdh
fjcb
ei

• d is the next root; g is in the left
subtree; h is in the right subtree.
a
gdh
fjcb
ei
a
g
fjcb
eid
h

Inorder And Postorder
• Scan postorder from right to left using
inorder to separate left and right subtrees.
• inorder = g d h b e i a f j c
• postorder = g h d i e b j f c a
• Tree root is a; gdhbei are in left subtree; fjc
are in right subtree.

Lecture 8 tree traversal

More Related Content

What's hot (10)

Similar to Lecture 8 tree traversal (12)

More from Abirami A (12)

Recently uploaded (20)

Lecture 8 tree traversal