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.

// 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;
    }
}