[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] :

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