LeetCode 156. Binary Tree Upside Down（树的倒转）

原题网址：https://leetcode.com/problems/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  

方法：递归。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    // 这个参数表示未来，未来的root，未来的left，未来的right，未来的flip函数啊，请帮我反转吧！！！
    private TreeNode flip(TreeNode fRoot, TreeNode fLeft, TreeNode fRight) {
        // 我现在就帮你反转，但请容我先记住必要的左子树
        TreeNode fl = fRoot.left;
        TreeNode fr = fRoot.right;
        fRoot.left = fLeft;
        fRoot.right = fRight;
        if (fl == null) return fRoot;
        return flip(fl, fr, fRoot);
    }
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if (root == null || root.left == null) return root;
        TreeNode fl = root.left;
        TreeNode fr = root.right;
        root.left = null;
        root.right = null;
        return flip(fl, fr, root);
    }
}