114. 二叉树展开为链表

class Solution {
public:
void flatten(TreeNode* root)
{
	if (root == NULL)
		return;
	TreeNode* now = root;
	TreeNode* pre;
	while (now)
	{
		if (now->left)
		{
			pre = now->left;
			while (pre->right)
			{
				pre = pre->right;
			}
			//当年节点的右节点挂在当前节点的左子树最右边的孩子上
			pre->right = now->right;
			//当前节点的左子树变为右子树，左子树置为NULL
			now->right = now->left;
			now->left = NULL;
		}
		now = now->right;
	}
}
};