The statement is False.

Reason:

A Binary Search Tree (BST) is a tree where for any given node:

• All values in the left subtree are less than the value of the node.
• All values in the right subtree are greater than the value of the node.

If we square all values in the BST, this property does not necessarily hold.

For example, consider the following BST:

    3
   / \
  1   4

Squaring all values gives:

    9
   / \
  1   16

After squaring, the left child (1) is still less than the root (9), but the right child (16) is greater than the root (9), so it still looks like a valid BST. However, consider a BST with negative values:

     0
    / \
  -3   2

Squaring all values results in:

    0
   / \
   9   4

After squaring:

• The left child (9) is no longer less than the root (0).

Thus, the tree is no longer a valid BST. Therefore, the given statement is False. Squaring the values in a BST does not necessarily preserve the BST property.

Would you like more details or have any questions? Here are 5 related questions:

1. What are the properties of a Binary Search Tree (BST)?
2. How does squaring affect the relative order of numbers?
3. What are other operations that do not preserve the BST structure?
4. Can you create a BST where squaring does preserve the BST property?
5. How can we prove that an operation preserves the BST property?

Tip: Always think about the properties that define a structure when applying transformations to see if they still hold!