[LeetCode每日一题]114. Flatten Binary Tree to Linked List

题目如下：

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

初看这道题觉得很简单，就是一个dfs，上手写起来后才发现没那么容易，这道题要求直接对原二叉树进行修改，开List遍历二叉树再将List中的节点连接的办法没能AC，看了讨论区以后发现下面这个解法比较容易理解。

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