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);    }}