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

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

1、對於當前訪問的node，假設左右兩子樹均為符合要求的展開連結串列

2、函式helper返回值為當前樹展開後的尾部node

3、根據左右子樹返回的連結串列尾部node，將當前父節點與左右子樹串聯成新的連結串列

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