Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int> > res; vector<int> temp; if(root!=nullptr) sumRec(res, temp, root, 0, sum); else return res; } void sumRec(vector<vector<int> > & res, vector<int>& temp ,TreeNode* root, int cur, const int sum) { cur += root->val; temp.push_back(root->val); if(root->left==nullptr && root->right==nullptr) { if(cur==sum) { res.push_back(temp); } } if(root->left != nullptr) { sumRec(res, temp, root->left, cur, sum); } if(root->right!=nullptr) { sumRec(res,temp,root->right, cur, sum); } temp.pop_back(); cur-=root->val; } };
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { // Start typing your C/C++ solution below // DO NOT write int main() function //dfs preorder vector<vector<int>> res; vector<int> temp; if(root==nullptr) return res; sumRec(res, temp, root, sum, 0); return res; } void sumRec(vector<vector<int>>& res, vector<int>& temp, TreeNode * root, const int sum, int cur) { cur += root->val; temp.push_back(root->val); if(root->left==nullptr && root->right==nullptr) { if(sum==cur) { res.push_back( temp); } return; } if(root->left!=nullptr) { sumRec(res, temp, root->left, sum, cur); temp.pop_back(); } if(root->right!=nullptr) { sumRec(res, temp, root->right, sum, cur); temp.pop_back(); } } };