LeetCode 156: 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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode UpsideDownBinaryTree(TreeNode root) {
        TreeNode node = root, parent = null, parentRight = null;
        
        while (node != null)
        {
            TreeNode left = node.left;
            node.left = parentRight; // Kernel statement
            parentRight = node.right;
            node.right = parent; // Kernel statement
            parent = node; // From top to down
            node = left; // Along with left side
        }
        
        return parent;
    }
}