156. Binary Tree Upside Down反转二叉树

［抄题］：

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:

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  

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

［英文数据结构或算法，为什么不用别的数据结构或算法］：

二叉树中用left/right是recursive，类似图中的公式

一个个节点去写是iterative，类似图中的for循环

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

六步里面：要提前把temp节点存起来，传递给cur.left

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

TreeNode类这样写 另外加一个item参数 相当于新建指针了
然后solution里要包括一个root

class TreeNode
{
  int data;
  TreeNode left, right;

  //parameter is another item
  TreeNode(int item) {
    data = item;
    left = right = null;
  }
}

新建节点需要tree.root = new，调用数据需要.data

class MyCode {
  public static void main (String[] args) {
    Solution tree = new Solution();
    tree.root = new TreeNode(1);
    tree.root.left = new TreeNode(2);
    tree.root.right = new TreeNode(3);
    tree.root.left.left = new TreeNode(4);
    tree.root.left.right = new TreeNode(5);

    TreeNode t = tree.upsideDownBinaryTree(tree.root);
    System.out.println("t.root.data = " + t.data);
    System.out.println("t.root.left.data = " + t.left.data);
  }

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

// package whatever; // don't place package name!

import java.io.*;
import java.util.*;
import java.lang.*;

class TreeNode
{
  int data;
  TreeNode left, right;

  //parameter is another item
  TreeNode(int item) {
    data = item;
    left = right = null;
  }
}

class Solution {
  //new root
    TreeNode root;
  //method, need parameter, there is return
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        //ini: prev, cur, next;
        TreeNode cur = root;
        TreeNode prev = null;
        TreeNode temp = null;
        TreeNode next = null;

        //iteration
        while (cur != null) {
            next = cur.left;
            cur.left = temp;
            temp = cur.right;
            cur.right = prev;

            prev = cur;
            cur = next;
        }

        return prev;
    }
}

class MyCode {
  public static void main (String[] args) {
    Solution tree = new Solution();
    tree.root = new TreeNode(1);
    tree.root.left = new TreeNode(2);
    tree.root.right = new TreeNode(3);
    tree.root.left.left = new TreeNode(4);
    tree.root.left.right = new TreeNode(5);

    TreeNode t = tree.upsideDownBinaryTree(tree.root);
    System.out.println("t.root.data = " + t.data);
    System.out.println("t.root.left.data = " + t.left.data);
  }
}

[潜台词] ：