LeetCode OJ：Flatten Binary Tree to Linked List（捋平二叉树）

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

也就是说用先序遍历的方式将所有的节点都放到右侧来，我这里的方法的主要思想就是每次左侧存在节点的时候就将其原封不动的搬移到右侧来，代码如下：

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     void flatten(TreeNode* root) {
         while(root){
             if(root->left){
                 TreeNode * leftBegin = root->left;
                 root->left = NULL;
                 TreeNode * leftEnd = leftBegin;
                 while(leftEnd->right)
                     leftEnd = leftEnd->right;
                 TreeNode * tmpRight = root->right;
                 root->right = leftBegin;
                 leftEnd->right = tmpRight;
             }
             root = root->right;
         }
     }
 };

用java写了一遍，方法与上面基本相同，代码如下：

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