5.4.4 Path Sum II

Notes: 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] ]  Solution: DFS.  */   /** * 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) {        vector<vector<int>> res;        vector<int> path;        pathSumRe(root, sum, res, path);        return res;    }    void pathSumRe(TreeNode *root, int sum, vector<vector<int>> &res, vector<int> &path)    {        if (!root) return;        path.push_back(root->val);        if (!root->left && !root->right && root->val == sum)        {            res.push_back(path);        }        pathSumRe(root->left, sum - root->val, res, path);        pathSumRe(root->right, sum - root->val, res, path);        path.pop_back();    }};