1008. 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.)

Example 1:

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

Note: 

1. 1 <= preorder.length <= 100
2. The values of preorder are distinct.

class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if(preorder == null) return null;
        int[] index = new int[1];
        return preOrderUtil(preorder,index, Integer.MAX_VALUE, Integer.MIN_VALUE);
    }
    public TreeNode preOrderUtil(int[] preorder,int[] index, int max, int min){
        int i = index[0];
        if(i>=preorder.length){
            return null;
        }
        int key = preorder[i];
        if(key > min && key < max){
            TreeNode node = new TreeNode(key);
            index[0]++;
            node.left = preOrderUtil(preorder, index, key, min);
            node.right = preOrderUtil(preorder, index, max, min);
            return node;
        }else{
            return null;
        }
    }
}