# Construct Binary Search Tree from Preorder Traversal

## Description

Return the root node of a binary **search** tree that matches the given `preorder` traversal.

_(Recall that a binary search tree is a binary tree where for every node, any descendant of `node.left` has a value `<`_ _`node.val`, and any descendant of `node.right` has a value `>`_ _`node.val`.  Also recall that a preorder traversal displays the value of the `node` first, then traverses `node.left`, then traverses `node.right`.)_

It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.

**Example 1:**

> **Input:** \[8,5,1,7,10,12]&#x20;
>
> **Output:** \[8,5,10,1,7,null,12]

![](https://3217298702-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-M5BYNv0H5fXhaR-4sW4%2F-M6QZtrM7qeA6PaMZPDX%2F-M6QbT7MFM_y3jc05-bX%2Fimage.png?alt=media\&token=95c8cf08-fbf2-40ea-ad93-bcedfe6b9989)

**Constraints:**

* `1 <= preorder.length <= 100`
* `1 <= preorder[i] <= 10^8`
* The values of `preorder` are distinct.

## **Code**

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int i = 0;
    public TreeNode bstFromPreorder(int[] A) {
        return bstFromPreorder(A, Integer.MAX_VALUE);
    }

    public TreeNode bstFromPreorder(int[] A, int bound) {
        if (i == A.length || A[i] > bound) return null;
        TreeNode root = new TreeNode(A[i++]);
        root.left = bstFromPreorder(A, root.val);
        root.right = bstFromPreorder(A, bound);
        return root;
    }
}
```
