Reconstruct the Binary Search Tree Described by Preorder

### Problem Given a valid preorder traversal of a binary search tree with distinct values, construct that tree and return its root. Every left descendant is strictly smaller than its ancestor and every right descendant strictly greater. Preorder visits root,left subtree,then right subtree. ```cpp class Solution { public: TreeNode* bstFromPreorder(vector<int>& preorder); }; ``` ### Examples **Example 1:** ```text Input: preorder = [6,2,1,4,9,8] Output: tree displayed as[6,2,9,1,4,8] Explanation: Values2,1,4 form the left subtree of6;9 with left child8 forms its right subtree. ``` **Example 2:** ```text Input: preorder = [3,2,1] Output: tree displayed as[3,2,null,1] Explanation: Each next value belongs in the preceding node's left subtree. ``` ### Constraints - `1 <= preorder.length <= 100`. - `1 <= preorder[i] <= 1000`. - All values are unique. - The input is guaranteed to be a valid BST preorder traversal. ### Contract considerations Preserve the original TreeNode pointer return and supplied node definition; arrays only display example shapes. ```hint Read the boundary condition The next preorder value belongs to a subtree only if it lies inside all inherited BST bounds. ``` ### Boundary reminders Preserve the original TreeNode pointer return and supplied node definition; arrays only display example shapes.