113. Path Sum II

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,

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool hasPathSum(struct TreeNode* root, int sum,vector<int> &myVec,vector<vector<int>> &listVec) {        if(root == NULL){            return false;        }        //this node is leaf node        if((!root->left)&&(!root->right)){            if(sum == root->val){                //push this vector;                myVec.push_back(root->val);                listVec.push_back(myVec);                myVec.pop_back();                return true;            }        }//leaf Node        if(root->left){            myVec.push_back(root->val);            hasPathSum(root->left,sum-root->val,myVec,listVec);            myVec.pop_back();        }        if(root->right){            myVec.push_back(root->val);            hasPathSum(root->right,sum-root->val,myVec,listVec);            myVec.pop_back();        }        return true;    }    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> result;        vector<int> myVec;        hasPathSum(root,sum,myVec,result);        return result;    }};