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# Construct Binary Search Tree from Preorder Traversal

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]
Output: [8,5,10,1,7,null,12]

```

**Constraints:**

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

```java
// Naive Solution -> O(N*N)
// Go to every node & find the split in the array and call recursively

// O(N) -> Solution
// Call left side first and maintain a global index, so that when it comes back to root node,
// it will be pointing to the split for right subtree.
// Now to maintain & check the BST order, we will mainatin upper & lower bounds.
class Solution {
    int index;

    public TreeNode bstFromPreorder(int[] preorder) {
        index = 0;
        return helper(preorder, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }

    public TreeNode helper(int[] preOrder, int lowerBound, int upperBound) {
        if (index >= preOrder.length || (preOrder[index] < lowerBound || preOrder[index] > upperBound))
            return null;

        TreeNode root = new TreeNode(preOrder[index++]);
        root.left = helper(preOrder, lowerBound, root.val - 1);
        root.right = helper(preOrder, root.val + 1, upperBound);

        return root;
    }
}
```


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