Path Sum II Solutions in C++

// Source : https://oj.leetcode.com/problems/path-sum-ii/
// Author : Hao Chen
// Date   : 2014-07-01



/**
 * 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> > result;
        vector<int> v;
        generatePathSum(root, sum, 0, v, result);
        return result;
    }

    void generatePathSum(TreeNode *root, int sum, int s, vector<int> v, vector<vector<int> >& result) {
        if (root==NULL) return ;

        s += root->val;
        v.push_back(root->val);

        if ( root->left==NULL && root->right==NULL) {
            if (s == sum) result.push_back(v);
            return;
        }

        generatePathSum(root->left, sum, s, v, result);
        generatePathSum(root->right, sum, s, v, result);
    }
};

#include <vector>
#include <cstddef>
#include <stack>
#include <algorithm>
using std::vector;

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>> ret;
        vector<int> path;
        pathSum(root, sum, path, ret);
        return ret;
    }
    void pathSum(TreeNode *root, int sum, vector<int> &path, vector<vector<int>> &ret) {
        if (!root) return;
        path.push_back(root->val);
        if (!root->left && !root->right && root->val == sum) ret.push_back(path);
        if (root->left) pathSum(root->left, sum-root->val, path, ret);
        if (root->right) pathSum(root->right, sum-root->val, path, ret);
        path.pop_back();
    }
};

/// Source : https://leetcode.com/problems/path-sum-ii/description/
/// Author : liuyubobobo
/// Time   : 2018-10-16

#include <iostream>
#include <vector>

using namespace std;


/// Recursion
/// Time Complexity: O(n)
/// Space Complexity: O(n)

/// 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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {

        vector<vector<int>> res;
        if(!root)
            return res;

        vector<int> tres;
        dfs(root, tres, 0, sum, res);
        return res;
    }

private:
    void dfs(TreeNode* node, vector<int>& tres, int tsum,
             int sum, vector<vector<int>>& res){

        tres.push_back(node->val);
        tsum += node->val;

        if(!node->left && !node->right){
            if(tsum == sum)
                res.push_back(tres);
        }
        else {
            if (node->left)
                dfs(node->left, tres, tsum, sum, res);

            if (node->right)
                dfs(node->right, tres, tsum, sum, res);
        }

        tres.pop_back();
        return;
    }
};


int main() {

    return 0;
}