［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] ：