LeetCode - Increasing Order Search Tree

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

 

Example 1:


Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
Example 2:


Input: root = [5,1,7]
Output: [1,null,5,null,7]
 

Constraints:

The number of nodes in the given tree will be in the range [1, 100].
0 <= Node.val <= 1000

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode increasingBST(TreeNode root) {
        Queue<Integer> queue = new LinkedList<>();
        helper(root, queue);
        TreeNode res = new TreeNode(queue.poll());
        TreeNode cur = res;
        while(!queue.isEmpty()) {
            cur.right = new TreeNode(queue.poll());
            cur = cur.right;
        }
        return res;
        
    }
    private void helper (TreeNode node, Queue<Integer> queue) {
        if (node == null) {
            return;
        }
        helper(node.left, queue);
        queue.add(node.val);
        helper(node.right, queue);
    }
    
    
}