This problem is very interesting. Recursive approach is very straightforward, though solution is likely of O(nlogn)/O(logn) time/space complexity (avg), or O(N^2)/O(N) worst case.

The key to optimize is to recall how iterative(non-recursive) pre-order traversal works, then follow the same idea to "reconstruct" BST from preorder sequence, during which we need a stack to keep track of node path from root. The trick here is to use input preorder sequence to simulate the stack in-place. For verification purpose, we need to maintain a "lower bound" variable: when we visit a node that is the right child of its parent, the parent's value must be lower bound of all subsequent values.

```
class Solution {
public:
bool verifyPreorder(vector<int>& preorder) {
if (preorder.empty()) return true;
int lowerBound = INT_MIN;
int stackTopPosition = 0;
for (int pos = 1; pos < preorder.size(); ++pos) {
if (preorder[pos] < lowerBound) return false;
while (stackTopPosition >= 0 && preorder[pos] > preorder[stackTopPosition]) {
lowerBound = preorder[stackTopPosition--]; // Update
}
preorder[++stackTopPosition] = preorder[pos]; // Push to stack
}
return true;
}
};
```