897. Increasing Order Search Tree

/*
// Given a 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 1 right child.

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

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode increasingBST(TreeNode root) {
        List<Integer> vals = new ArrayList();
        inorder(root, vals);
        TreeNode ans = new TreeNode(0);
        TreeNode cur = ans;
        for (int v: vals) {
            cur.right = new TreeNode(v);
            cur = cur.right;
        }
        return ans.right;
    }

    public void inorder(TreeNode node, List<Integer> vals) {
        if (node == null) {
            return;
        }
        
        inorder(node.left, vals);
        vals.add(node.val);
        inorder(node.right, vals);
    }
}