construct-binary-search-tree-from-preorder-traversal.java

// Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.
//
// It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.
//
// A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.
//
// A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.
//
//  
// Example 1:
//
//
// Input: preorder = [8,5,1,7,10,12]
// Output: [8,5,10,1,7,null,12]
//
//
// Example 2:
//
//
// Input: preorder = [1,3]
// Output: [1,null,3]
//
//
//  
// Constraints:
//
//
// 	1 <= preorder.length <= 100
// 	1 <= preorder[i] <= 1000
// 	All the values of preorder are unique.
//
//


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int i;
    public TreeNode bstFromPreorder(int[] preorder) {
        return helper(preorder, Integer.MAX_VALUE);
    }
    public TreeNode helper(int[] preorder, int n) {
        if (i == preorder.length || preorder[i] > n) return null;
        TreeNode node = new TreeNode(preorder[i++]);
        node.left = helper(preorder, node.val);
        node.right = helper(preorder, n);
        return node;
    }
}