Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

每個節點的右節點都是preorder traverse的下一個節點 -> 應該先 反preorder遍歷 (即node.right->node.left->node) 到最後，從最後開始relink

time: O(n), space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode prev = null;
    
    public void flatten(TreeNode root) {
        
 
if(root == null) {
            return;
        }
        
        flatten(root.right);
        flatten(root.left);
        
        root.right = prev;
        root.left = null;
        prev = root;
    }
}