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转换树需要多少个右旋？

[英]How many Right Rotation is need to convert a tree?

原文 2015-03-26 10:55:24 5 1 algorithm/ data-structures/ graph/ tree/ graph-theory

I select one nice interview question. 我选择一个不错的面试问题。 Can anyone clarify it for me? 谁能为我澄清一下？

Suppose a binary tree with 6 nodes is given, such that each node has only left childs. 假设给出了一个具有6个节点的二叉树，因此每个节点只有左子节点。 With how many "right rotate" operations (without any left rotates), we can convert this tree to a tree in which each node has only right childs? 通过执行多少次“向右旋转”操作（没有任何向左旋转），我们可以将该树转换为每个节点仅具有右子节点的树吗？

my solution: 我的解决方案：

I think n-1 rotation is enough (by simulation), but I‌ couldent make a proof, which expert could help me for proof or idea? 我认为n-1旋转就足够了（通过模拟），但是我可以做一个证明，哪位专家可以帮助我进行证明或提出想法？

1 个解决方案

The tree starts with a root node and 5 children (which is n-1 children) to the left. 该树从root开始，左侧有5个子节点（ n-1个子节点）。 Each rotation around the root increases the number of children on the right by 1. So after 5 rotations (meaning n-1 rotations) all of the children will be on the right. 围绕root每次旋转都会使右侧的子root增加1。因此，在旋转5圈（即n-1旋转）之后，所有子代都将在右侧。

Proof by induction: propose that after n rotations there are n children on the right. 通过归纳证明：提出n旋转后，右边有n孩子。

Step 1: After 1 rotation there is 1 child on the right. 步骤1：旋转1次后，右边有1个孩子。

Step 2: Assume that after n rotations there are n children on the right, and prove that after n+1 rotations, there are n+1 children on the right. 步骤2：假设在n旋转之后，右边有n孩子，并证明在n+1旋转之后，右边有n+1孩子。

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