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.

Input: [1,2,3,4,5]

    1

   / \

  2   3

 / \

4   5

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

   4

  / \

 5   2

    / \

   3   1 

Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1

  / \

 2   3

    /

   4

    \

     5

The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].

题意：

将一棵树，按照一种方式上下颠倒

Solution1： Tree + Recursion

Need save the tree information before changing the tree structure

Now, the basic idea is to go though from the original root (2), make root’s left child(4)  as newParent whose left child will be the original root’s right child (5) and right child will be the original root (2). After that, you set new root to root.left (5), and process it in the same way.

代码：

 class Solution {

     public TreeNode upsideDownBinaryTree(TreeNode root) {

         if(root == null || root.left == null) return root;

         TreeNode newNode = upsideDownBinaryTree(root.left); // newNode最后要返回，所以找到、存下来后，就不再动了

         root.left.left = root.right;// 这里为何不是newNode.left = root.left ?

         root.left.right = root;

         root.left = null ;

         root.right = null;

         return newNode;

     }

 }