Validate Binary Search Tree

A binary tree is represented by aligned arrays. Node i has value values[i]; left[i] and right[i] contain child indices, or -1 when that child is absent. The index root identifies the root node.
Return true if the tree is a valid binary search tree and false otherwise.
For every node, every value in its left subtree must be strictly smaller and every value in its right subtree must be strictly greater. Therefore duplicate values make the tree invalid.

Function
isValidBST(values: int[], left: int[], right: int[], root: int) → boolean

Examples
Example 1
values = [2,1,3]
left = [1,-1,-1]
right = [2,-1,-1]
root = 0
return = true
The left value 1 is smaller than 2, and the right value 3 is greater.
Example 2
values = [5,1,4,3,6]
left = [1,-1,3,-1,-1]
right = [2,-1,4,-1,-1]
root = 0
return = false
Node value 4 lies in the right subtree of 5 but is not greater than 5.

The trick is to pass a valid range down the tree. Start at root with low = -infinity and high = +infinity. For node i, require low < values[i] < high. Go left with high = values[i]. Go right with low = values[i]. Strict comparisons on both sides handle the duplicate rule for free. The pitfall is comparing a node only to its parent. In Example 2, node 4 is greater than nothing it's compared to locally, since its children are fine, but it sits in the right subtree of 5 and is less than 5. A parent-only check passes it. Use long or null bounds so values at the int limits don't break the sentinel. An iterative stack or an in-order traversal that checks strictly increasing order also works. Recursion depth can hurt on a skewed tree, so mention the iterative option. If you freeze live, StealthCoder is the hedge, but the range-passing idea fits in five lines.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.