A BST can be broken by accessing one of its nodes, how can I always make sure this happens? I have an assignment that asks for this. So I am not looking for the answer itself but a hint on how to find a value that will always break the BST condition by myself.

If I have access to any node N of a BST that has at least 2 nodes in its root, and where every node will always be different.

If I were to replace N.key with x, what should the value of x be in order for the tree not to be a BST anymore?

I can use the following operations:
GetPre = key before N when seen from an inorder transverse.

GetSuc = key after N when seen from an inorder transverse.

MAX = gets the node value from the sub-tree that has N as root.

MIN = gets node value from the sub-tree that has N as root.

I broke it in several subcases, but I am sure this can be written as one single formula.

I came up with:

$x = N.key - GetPre(N)$ but it doesn't work as I expected.

I appreciate any insight.
Hint: In a BST, an inorder traversal produces the keys in sorted order. If we had N.key = $20$ and the inorder traversal was $2, 4, 10, 11, 17, 20, 28, 31, 40$, what $x$ values would break the sorted order when put in the place of $20$?