LintCode_453 将二叉树拆成链表

题目

将一棵二叉树按照前序遍历拆解成为一个假链表。所谓的假链表是说，用二叉树的 right 指针，来表示链表中的 next 指针。

样例

              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;
		}
	}
}