So we can start b-search on string length of s1, for s2, since current substr len is fixed, we use rolling-hash to check match. Also, another interesting point to handle collision in a typical BK algorithm: we use double hashing to avoid collision, instead of a O(n) brutal-force str...
LintCode "Remove Node in Binary Search Tree" <2025年4月> 日一二三四五六 303112345 6789101112 13141516171819 20212223242526 27282930123 45678910 Not hard to find a solution, but there are several corner cases. View Code
Is it a binary tree Questions please seeHackerRank. It was too long. In simple, how can you check if a tree is a binary search tree? Idea Read the binary search tree from leftmost to rightmost, and make sure the order is always increasing. If you still remember how to do an in-orde...
Given an array where elements are sorted in ascending order, convert it to a height balanced BST. Accepted Code: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* ...
classSolution {public:/** *@param n: Given a decimal number that is passed in as a string *@return: A string*/stringbinaryRepresentation(stringn) { size_t pos= n.find("."); unsignedlongvi = atoi(n.substr(0, pos).c_str());doublevf =atof(n.substr(pos).c_str());//Int part...
原题地址 Unique Binary Search Trees(参见这篇文章)的升级版 做题的时候我在想,这要是把每个二叉树都独立创建一份得多麻烦啊,试试能不能共用"公共部分",试了一下,果然可以,哈哈。 trees[i][j]表示数字i到j所能组成的所有二叉树的根节点 代码:
Tony's Log 2025年4月> 日一二三四五六 303112345 6789101112 13141516171819 20212223242526 27282930123 45678910 Recursion, a natural thought: View Code Iterative, just picture it in your mind :) View Code
HM r=rl;for(auto &kv : rr) r[kv.first] +=kv.second; r[p->val] ++;returnr; }public: vector<int> findMode(TreeNode*root) { auto h=go(root);intmcnt =0; vector<int>ret;for(auto &kv : h) {if(kv.second >mcnt)
23 This is to test your knowledge on BST and its traversal. Flatting BST into an array using in-order, and check that array. It is that simple: classSolution {public:voidserial(TreeNode *root, vector<int> &vec) {if(!root)return;if(root->left) serial(root->left, vec); ...