One TopCoder article introduces a very interesting alternative solution to Longest Common Sequences problem. It is based on this statement (http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=stringSearching): The important point that allows us to use BS is the fact that if the gi...
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* };9*/10classSolution {11public:12TreeNode *sortedArrayToBST(vector<int> &num,intstart,intend) {13if(start >en...
Not hard to find a solution, but there are several corner cases. View Code
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...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution { vector<vector<int>>ret;intgo(TreeNode *p)//return max height{if(!p)return0;inthl = ...
12 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); ...
It is a matter of hashmap use. For the follow-up: typedef unordered_map<int, unsigned>HM;classSolution { HM go(TreeNode*p) {if(!p)returnHM(); auto rl= go(p->left); auto rr= go(p->right); HM r=rl;for(auto &kv : rr) r[kv.first] +=kv.second; ...
Here is a BFS solution which is compatible with LeetCode binary tree format.class Solution {public: /** * This method will be invoked first, yo...
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-order recursive visit for ...