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  

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization: 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}".

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if (root == null || root.left == null) return root;
        TreeNode oldLeft = root.left;
        TreeNode oldRight = root.right;
        TreeNode newRoot = upsideDownBinaryTree(oldLeft);
     
        oldLeft.left = oldRight;
        oldLeft.right = root;
     
        root.left = null;
        root.right = null;
        return newRoot;
    }
}

========
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        TreeNode pre = null;
        TreeNode cur = root;
        TreeNode oldRight = null;
        while (cur != null) {
            TreeNode next = cur.left;
            cur.left = oldRight;
            oldRight = cur.right;
            cur.right = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}