8月刷题/Mock interview接龙活动（缺米刷题的来）

第一天打卡：
serialize and deserialize binary tree: node + left_subtree + right_subtree
k closest points to origin: quick select/heap

8.13  Q692,1046, 703  pq的sorting怎么这么乱。

【8.13打卡】 354，48，54，59，73，329
今天练习了矩阵的题目，用到的算法思路挺多挺高级，考虑的情况比较复杂，需要多琢磨几下，明天继续矩阵。。。

8.13 day13 August LeetCoding Challenge

打卡第39天 2020.08.13
501. 二叉搜索树中的众数
559. N叉树的最大深度
530. 二叉搜索树的最小绝对差
783. 二叉搜索树节点最小距离
563. 二叉树的坡度
589. N叉树的前序遍历
590. N叉树的后序遍历
429. N叉树的层序遍历

8月12日 刷题第十二天 打卡十二天

今天复习了老题
1. Fibonacci
-DP 做
-2个int variables
2. A to the power of B
-edge cases & corner case check
3. Debug A to the power of B
-debug -> a == 1;
4. Quick Sort
5. MergeSort
6. Rainbow sort;

两道之前未做过的新题：
1. find the closest Node in BST
2. Find K closest Nodes in BST
Approach one:
use inOrder traversal and put it into the array then use binary Search

Approach two:
       //High level:
       //use a maxHeap to store the K closest elements of target
       //Elements in the maxHeap are sorted based
       //on the gap between themselves and target, in another word,
       //the top element in the maxHeap has the largest
       //gap between itself and target;
       //Details:
       //if the size of maxHeap is smaller than K, put the current element into the maxHeap;
       //else if the gap is smaller than the top one in the maxHeap, poll it out and put
       //the element into it;
       //use in-Order traversal iteratively search the tree and
       //maintain the size == k maxHeap (if k <= nodes in the given root)
       //put all elements into the maxHeap(if k > nodes in the given root)
       //T:O(n);
       //S:O(k + logk) -> O(k);

Approach three:
//use a list to store k elements which are currently the k closest nodes with target
then keep traverse the remaining nodes of the tree and dynamically maintain the list;
//until we traversed all elements in the tree, shall we return;

加油加油！

Day 51, solved 1 problem.

8.13打卡，leetcode 43, 625,5, 53

08/13 三题

29        Divide Two Integers    这题真的是很惭愧，错了5次，注意要预先处理edge cases，被除数小于除数，除数等于1，当被除数为最小的负整数时会溢出，利用binary search的原理计算被除数里包含了几个除数，递归
78        Subsets   从空集开始，每遇到一个新数字就加到原有的集合里，记得先把原有的集合复制一遍
140  Word Break II    遍历字典，if S starts with a word, 拿出来这个word，把S的substring再用字典遍历一次，递归。优化：用HashMap记忆找过的字符串，预先计算字典里word的最大长度和最小长度

8.13打卡 7道bitwise operation