Is it always possible to construct the unique binary tree from the preorder and Postorder?

It is not possible to construct a general Binary Tree from preorder and postorder traversals (See this).

Can you find a unique tree when any two traversals are given?

It depends on what traversals are given. If one of the traversal methods is Inorder then the tree can be constructed, otherwise not. Therefore, following combination can uniquely identify a tree.

Can we have pre-order and post-order traversal of a binary tree same?

With the above example, pre-order will produce AB or AC respectively and post-order will produce BA and CA. Thank you, So just in one situation the pre-order and post-order are the same.

Is it possible to construct a binary tree uniquely?

We can’t construct a unique binary tree from its level-order & postorder traversals. So, we can’t construct the tree uniquely.

Is preorder traversal unique?

Preorder and postorder do not uniquely define a binary tree. 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.

What is common in three different types of traversals inorder preorder and Postorder?

What is common in three different types of traversals( inorder,preorder and postorder)

• Root is visited before right subtree.
• Left subtree is always visited before right subtree.
• Root is visited after left subtree.
• All of the above.

Is post Order opposite of preorder?

Reason is post order is non-tail recursive ( The statements execute after the recursive call). If you just observe here, postorder traversal is just reverse of preorder traversal (1 3 7 6 2 5 4 if we traverse the right node first and then left node.)

Where is preorder traversal from binary tree?

Preorder Traversal: Preorder traversal will create a copy of the tree….Algorithm for binary tree traversal

1. Traverse the left sub-tree, (recursively call inorder(root -> left).
2. Visit and print the root node.
3. Traverse the right sub-tree, (recursively call inorder(root -> right).