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1. BINARY TREE CONSTRUCTION

2. Binary Tree Construction using given tree traversal
 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

3. Some Examples
preorder
= ab
a
b
a
b
inorder
= ab
b
a
a
b
postorde
r = ab
b
a
b
a

4. Binary Tree Construction using given tree traversal
 Can you construct the binary tree, given two
traversal sequences?
 Depends on which two sequences are given
 Inorder And Preorder
 Inorder And Postorder

5. Binary Tree Construction using given tree traversal
 By given preorder and postorder we will come to
know which is the root node.
 Ones we learn about the root node we can learn
about the LST (Left SubTree) and RST (Right
SubTree)
 Repeat these 2 steps until the end.

6. Example
 Given preorder: ABCDE
 Given Inoder: BADCE
 Step1: By given preorder we can identify A is root
node.
 Step 2: Find that root node in given inorder sequence.
BADCE

7. Cont…
 Result of above steps are:
 Next root node is B but there is no left and right child
 Repeat step 1 and 2 again

8. Cont….

9. Cont…

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

11. Inorder And Preorder
 preorder = a b d g h e i c f j
 inorder = g d h b e i a f j c
 Scan the preorder left to right
using the inorder to separate
left and right subtrees
1. g is the next root;
NULL are in the left subtree;
NULL are in the right subtree
2. h is the next root;
NULL is in the left subtree;
NULL is in the right subtree
3. e is the next root;
NULL is in the left subtree;
i is in the right subtree
a
g
fjc
b
ei
d
h
a
g
fjc
b
d
h
e
i

12. Inorder And Preorder
 preorder = a b d g h e i c f j
 inorder = g d h b e i a f j c
 Scan the preorder left to right
using the inorder to separate
left and right subtrees
1. i is the next root;
NULL are in the left subtree;
NULL are in the right subtree
2. c is the next root;
J F is in the left subtree;
NULL is in the right subtree
3. f is the next root;
NULL is in the left subtree;
j is in the right subtree
a
g
fjc
b
d
h
e
i
a
g
b
d
h
e
i
c
fj

13. Inorder And Preorder
a
g
b
d
h
e
i
c
f
j
Final Inorder Binary Tree From given Inorder and Preorder
traversal