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.)

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.

class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode root = new TreeNode(preorder[0]);
        stack.push(root);
        for (int i = 1; i < preorder.length; i++) {
            TreeNode node = new TreeNode(preorder[i]);
            if (preorder[i] < stack.peek().val) {                
                stack.peek().left = node;                
            } else {
                TreeNode parent = stack.peek();
                while (!stack.isEmpty() && preorder[i] > stack.peek().val) {
                    parent = stack.pop();
                }
                parent.right = node;
            }
            stack.push(node);            
        }
        return root;
    }
}