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 / / 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: void DFS(TreeNode *root, vector<int> &temp, int tempSum) { if(root->left == NULL && root -> right== NULL && tempSum == sum) { result.push_back(temp); return; } if(root->left!= NULL ){ temp.push_back(root->left->val); DFS(root->left, temp,root->left->val + tempSum ); temp.pop_back(); } if(root->right != NULL ){ temp.push_back(root->right->val); DFS(root->right, temp, root->right->val + tempSum); temp.pop_back(); } } vector<vector<int> > pathSum(TreeNode *root, int sum) { // Start typing your C/C++ solution below // DO NOT write int main() function result.clear(); if(!root) return result; this->sum = sum; vector<int> temp; temp.push_back(root->val); DFS(root, temp, root->val); return result; } private: int sum; vector<vector<int>> result; };