leetcode156- Binary Tree Upside Down- medium

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  

递归。主要操作指针。

1.左边也做一次。

2.现在的三个结点指针变一变。记得要把root的指针清null，不然会出现双向指针。

3.把左边指针传上来的root当root传回去，这样才能把最底下弄到的新root给传到最上面来。

实现：

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