Path Sum,path Sum II

Path SumTotal accepted:81706 Total submissions:269391 difficulty:easy

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such this adding up all the values along the Path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             /             4   8           /   /           /  4         /  \              7    2      1

Return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

/** Definition for a binary tree node. * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * TreeNode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public:    BOOLHaspathsum (treenode* root,intSumintcursum) {        if(!root)return false; Cursum+ = Root->Val; if(Root->left==null && root->right==NULL) {cout<<curSum<<Endl; if(cursum = = SUM)return true; return false; }        returnHaspathsum (root->left,sum,cursum) | | Haspathsum (root->right,sum,cursum); }    BOOLHaspathsum (treenode* root,intsum) {        returnHaspathsum (Root,sum,0); }};

Next challenges: (m) path Sum II (H) Binary Tree Maximum Path sum (m) sum Root to Leaf Numbers Path Sum IITotal accepted:65371 Total submissions:240118 difficulty:medium

Given a binary tree and a sum, find all root-to-leaf paths where each path ' s sum equals the Given sum.

For example:
Given the below binary tree and sum = 22,

              5             /             4   8           /   /           /  4         /  \    /         7 2 5   1

Return

[   [5,4,11,2],   [5,8,4,5]]

/** Definition for a binary tree node. * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right ; * TreeNode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution {Private:    voidPathsum (treenode* root, vector<vector<int>>& Res, vector<int>& Oneres,intSumintcursum) {        if(!root)return; Cursum+ = Root->Val; Oneres.push_back (Root-val); if(!root->left &&!root->Right ) {            if(Cursum = =sum)            {Res.push_back (oneres);            } oneres.pop_back (); return; } pathsum (Root-left,res,oneres,sum,cursum); Pathsum (Root-right,res,oneres,sum,cursum);    Oneres.pop_back (); } Public: Vector<vector<int>> pathsum (treenode* root,intsum) {Vector<vector<int>>Res; Vector<int>Oneres; Pathsum (Root,res,oneres,sum,0); returnRes; }};

Path Sum,path Sum II