原题地址：http://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/

题意：

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
Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

由上面可以看出：这道题的意思是将一颗二叉树平化（flatten）为一条链表，而链表的顺序为二叉树的先序遍历。

解题思路：首先将左右子树分别平化为链表，这两条链表的顺序分别为左子树的先序遍历和右子树的先序遍历。然后将左子树链表插入到根节点和右子树链表之间，就可以了。左右子树的平化则使用递归实现。

代码：

# Definition for a  binary tree node

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # @param root, a tree node

    # @return nothing, do it in place

    def flatten(self, root):

        if root == None:

            return

        self.flatten(root.left)

        self.flatten(root.right)

        p = root

        if p.left == None:

            return

        p = p.left

        while p.right:

            p = p.right

        p.right = root.right

        root.right = root.left

        root.left = None