Count invalid nodes in a BST

You are given the root of a binary tree. A valid Binary Search Tree (BST) must satisfy: for every node with value `v`, **all** values in its left subtree are strictly `< v`, and **all** values in its right subtree are strictly `> v` (i.e., the constraint is with respect to all ancestors, not just the parent). Instead of returning a boolean, return the **number of nodes that violate the BST rule**. ### Input - The root of a binary tree. - Node values are integers. ### Output - An integer: the count of nodes that are part of at least one BST-order violation (i.e., nodes whose value is not within the valid `(low, high)` range implied by ancestors). ### Notes / Clarifications - Use strict inequality (no duplicates allowed in a valid BST). - Counting is per node: if a node violates constraints from multiple ancestors, it is still counted **once**. ### Constraints (typical interview assumptions) - Number of nodes `n` up to ~`10^5`. - Aim for `O(n)` time.