LintCode_453 将二叉树拆成链表

              1
                    1          2
    / \             2   5    =>    3
  / \   \           3   4   6          4
                                           5
                                               6
思路
left = root->left;
right = root->right;
p = left->right的最后一个节点
如果有left节点
{
　　root->left = NULL;
　　root->right = left;
　　left->right = right;
}
然后递归
C++代码

void flatten(TreeNode *root) {
	// write your code here
	if (root)
	{
		TreeNode* left = root->left;
		TreeNode* right = root->right;
		TreeNode* p = root;
		flatten(root->left);
		flatten(root->right);
		if (left)
		{
			p = root->left;
			while (p->right)
			{
				p = p->right;
			}
			root->left = NULL;
			root->right = left;
			p->right = right;
		}
	}
}