hackerrank Diameter of Binary Tree 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 publicclassDiameterOfTree { publicstaticintdiameter =0; publicstaticintgetDiameter(BinaryTreeNode root) { if(root !=null) { intleftCount = getDiameter(root.getLeft()); intrightCount = getDiameter(roo...
25 26classSolution{public: vector<vector<int>>levelOrderBottom(TreeNode* root) {if(!root)returnvector<vector<int>>(); vector<vector<int> > res; queue<TreeNode*> q; q.push(root);while(!q.empty()) { vector<int> level;intsize = q.size();for(inti =0; i < size; ++i) { TreeN...
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...
* 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 = go(p->left);inthr = go(p->right);inth = max(hl, hr) +1;if(h...
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* };9*/10classSolution {11public:12TreeNode *sortedArrayToBST(vector<int> &num,intstart,intend) {...
Not hard to find a solution, but there are several corner cases.class Solution {public: /** * @param root: The root of the binary search tree. ...
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...
LintCode "Binary Tree Serialization" 2025年5月> 日一二三四五六 27282930123 45678910 11121314151617 18192021222324 25262728293031 1234567 Here is a BFS solution which is compatible with LeetCode binary tree format. View Code
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); ...
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; r[p->val] ++;returnr; ...