1008. Construct Binary Search Tree from Preorder Traversal

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]
          8
    5           10
1       7   null   12

Note:

1. 1 <= preorder.length <= 100

2. The values of preorder are distinct.

My Solutions:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) return null;
        TreeNode root = new TreeNode(preorder[0]);
        for (int i = 1; i < preorder.length; i++) {
            root = helper(root, preorder[i]);
        }       
        return root;
    }
    
    public TreeNode helper(TreeNode root, int val) {
        if (root == null) return new TreeNode(val);
        if (val < root.val) root.left = helper(root.left, val);
        else root.right = helper(root.right, val);
        return root;
    }
}