Binary Tree Upside Down

https://oj.leetcode.com/problems/binary-tree-upside-down/

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1   / \  2   3 / \4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4  / \ 5   2    / \   3   1  

 public TreeNode UpsideDownBinaryTree(TreeNode root)

这一题不是很难，原因就在与它的特殊的树的结构。也就是所有节点的右子树都只有一个节点。也就是实际上没有右子树的概念，只有右子节点。那么循环的做法就是从根节点不停往左节点往下遍历，将parent节点做为当前节点的右节点，兄弟节点作为当前节点的左节点。所以在每一层，你所需要保存的就是parent节点，兄弟节点（parent的右节点），以及下一层的左子节点。根据上述算法，给出解法如下：

 

    public TreeNode UpsideDownBinaryTree(TreeNode root) {        if(root == null)            return null;        TreeNode prev = root, neighbor = root.right, cur = root.left;//cur用来向左遍历，prev表示parent节点，neighbor表示兄弟节点        while(cur != null){            TreeNode cur_neighbor = cur.right, next = cur.left;//我们需要再改变树结构之前先行记录下一层的节点已经兄弟节点。            cur.right = prev;            cur.left = neighbor;            prev = cur;            neighbor = cur_neighbor;            cur = next;        }        root.left = null;        root.right = null;        return prev;    }}