# Can we construct Binary Tree from inorder or preorder or postorder (1 traversal only)?

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Yes, we can construct a Binary Tree from one traversal only (Inorder or Postorder or Preorder) but we cannot construct an unique Binary Tree from a single traversal (inorder or preorder or postorder). This is because from a given traversal, we do not have the complete information. We have the following information:

- Inorder -> We know the leftmost and rightmost node in Binary Tree
- Preorder -> We know the root node of the Binary Tree for sure
- Postorder -> We know the root node of the Binary Tree for sure

The three traversal in summary:

- Inorder: left, root, right
- Preorder: root, left, right
- Postorder: left, right, root

The problem is as follows:

# Problem with Inorder

With Inorder traversal, we know the leftmost and rightmost node in the Binary Tree but we do not know the root of the Binary Tree.

Hence, any node (including the leftmost or the rightmost node) can be the root of the Binary Tree.

If we consider the leftmost node as the root node, then we get a right skewed Binary Tree.

If we consider the rightmost node as the root node, then we get a left skewed Binary Tree.

The problem exists from every subtree of the Binary Tree. We have no information about the root node.

# Problem with Preorder

With Preorder traversal, we know the root of the Binary Tree for sure but we do not know the leftmost or rightmost node of the Binary Tree.

Consider the case when we can have no right sub-tree from the root. Then, the traversal will go into the leftsubtree directly and the last node will be in the left sub-tree and is not the rightmost node.

Therefore, we cannot create a unique Binary Tree using Preorder.

# Problem with Postorder

With Postorder traversal, we know the root of the Binary Tree for sure but we do not know the leftmost or rightmost node of the Binary Tree.

Consider the case when we can have no left sub-tree from the root. Then, the traversal will go into the right sub-tree directly and the first node will be in the right sub-tree and is not the leftmost node.

Therefore, we cannot create a unique Binary Tree using Postorder.

# Which combinations will work?

To construct a unique Binary Tree, we need to know the exact nodes that are:

- Root
- Leftmost node
- Rightmost node

Preorder and Postorder traversal give us information about the root node.

Inorder traversal give us information about the leftmost and rightmost node.

Hence, if we take any two traversals: one being Inorder and the order from Preorder or Postorder, we can create a unique Binary Tree.

Note: We cannot create a Unique Binary Tree from the combination of Preorder and Postorder traversal as we have the information of the root only.

Hence, combinations from which we can create a unique Binary Tree are:

- Inorder + Postorder
- Inorder + Preorder

Combinations from which we cannot create a unique Binary Tree are:

- Inorder
- Postorder
- Preorder
- Preorder + Postorder

This gives you a good idea of whether it is possible to construct a Binary Tree from a given traversal like Inorder or Postorder or Preorder.