Given the in-order, and pre-order or post-order sequences of a binary tree, how do we construct a binary tree from them?

Given only pre-order and post order traversals, we can not construct a unique binary tree from them.

For example given

In-order : {1, 2, 3, 4, 5, 6, 7, 8}

Pre-order: {5, 3, 2, 1, 4, 8, 6, 7}

We have to create the following binary tree

We use recursive approach for constructing a binary tree from In-order and Post-order or Pre-order traversals .

In Pre order, we first traverse the root, then left sub tree and right sub tree.

For example consider the following binary tree

1

/ \

2 3

/ \ / \

4 5 6 7

Inorder: {4, 2, 5, 1, 6, 3, 7}

Preorder: {1, 2, 4, 5, 3, 6, 7}

Create 1 as the root node,

We can use the index of 1 in Inorder (in this case 3) to calculate the size of the left and right subtrees.

Size(left subtree) = index

Size(right subtree) = size-index-1

We can also use this index to calculate the offset of Inorder and Post order sequences for the subtree.

To create the left subtree we pass Inorder:{4, 2, 5} Preorder:{2, 4, 5} as parameters to the recursive call.

To create the right subtree we pass Inorder:{6, 3, 7} Preorder:{3, 6, 7} as parameters.

In post order, we first traverse the left subtree, right subtree and visit the root.

Similar to the above algorithm, this element divides the in-order sequence into left subtree and right subtree.

Considering the same example as above the Post order is {4, 5, 2, 6, 7, 3, 1}

Once we create the root with the last element i.e 1.

We pass the following parameters to create the left and right subtrees.

Inorder Postorder

Left Subtree {4, 2, 5} {4, 5, 2}

Right Subtree {6, 3, 7} {6, 7, 3}

Here is the C++ implementation of the above two algorithms.

Given only pre-order and post order traversals, we can not construct a unique binary tree from them.

For example given

In-order : {1, 2, 3, 4, 5, 6, 7, 8}

Pre-order: {5, 3, 2, 1, 4, 8, 6, 7}

We have to create the following binary tree

We use recursive approach for constructing a binary tree from In-order and Post-order or Pre-order traversals .

**Constructing from Inorder and Preorder:**In Pre order, we first traverse the root, then left sub tree and right sub tree.

*So the first element in the pre order is always the root*. We search for this element in the in-order sequence. This index divides the in-order sequence into left subtree and right subtree.For example consider the following binary tree

1

/ \

2 3

/ \ / \

4 5 6 7

Inorder: {4, 2, 5, 1, 6, 3, 7}

Preorder: {1, 2, 4, 5, 3, 6, 7}

Create 1 as the root node,

- all the left elements to the left of it {4, 2, 5} form the left subtree
- all the elements to the right of it {6, 3, 7} form the right subtree

We can use the index of 1 in Inorder (in this case 3) to calculate the size of the left and right subtrees.

Size(left subtree) = index

Size(right subtree) = size-index-1

We can also use this index to calculate the offset of Inorder and Post order sequences for the subtree.

To create the left subtree we pass Inorder:{4, 2, 5} Preorder:{2, 4, 5} as parameters to the recursive call.

To create the right subtree we pass Inorder:{6, 3, 7} Preorder:{3, 6, 7} as parameters.

**Constructing from Inorder and Postorder:**In post order, we first traverse the left subtree, right subtree and visit the root.

*So the last element in the post order sequence is the root*.Similar to the above algorithm, this element divides the in-order sequence into left subtree and right subtree.

Considering the same example as above the Post order is {4, 5, 2, 6, 7, 3, 1}

Once we create the root with the last element i.e 1.

We pass the following parameters to create the left and right subtrees.

Inorder Postorder

Left Subtree {4, 2, 5} {4, 5, 2}

Right Subtree {6, 3, 7} {6, 7, 3}

Here is the C++ implementation of the above two algorithms.