Flatten Binary Tree to Linked List

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

For example,
Given

         1
        /        2   5
      / \        3   4   6

The flattened tree should look like:

   1
         2
             3
                 4
                     5
                         6

SOLUTION 1:使用递归解决，根据left是否为空，先连接left tree, 然后再连接右子树。使用一个tail 来记录链的结尾。在递归之前，先将root.left,root.right保存下来。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public void flatten(TreeNode root) {
12         dfs(root);
13     }
14
15     // return : the tail of the list.
16     public TreeNode dfs(TreeNode root) {
17         if (root == null) {
18             return null;
19         }
20
21         TreeNode left = root.left;
22         TreeNode right = root.right;
23
24         // Init the root.
25         root.left = null;
26         root.right = null;
27
28         TreeNode tail = root;
29
30         // connect the left tree.
31         if (left != null) {
32             tail.right = left;
33             tail = dfs(left);
34         }
35
36         // connect the right tree.
37         if (right != null) {
38             tail.right = right;
39             tail = dfs(right);
40         }
41
42         return tail;
43     }
44 }