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.

Example:

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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

实际上是把整个树🌳顺时针翻转了; 4 2 5的位置看出翻转的操作应该是什么样的.

我觉得还是要见过这种题，不然自己写迭代的那个解法实际上很容易错

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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 = null; // just for this time
    root.right = null;
    
    return newRoot;
}
  1. root是null或者是leaf node(root.left == null && root.right == null)的时候，就return；
  2. 每次翻转三角形三个节点的时候，右子节点暂时两个子节点都赋值为null;