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 Solution in Java: class Solution { public TreeNode bstFromPreorder(int[] preorder) { if(preorder == null || preorder.length == 0){ ...