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,

              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<int> path;
    vector<vector<int>> matrix;
    
    void pathSum3(TreeNode * root, int sum){
        
        stack<TreeNode *> s;
        
        int path_sum = 0;
        TreeNode * p = root;
        while(p || !s.empty()){
            while(p){
                s.push(p);
                
                path_sum += p->val;
                path.push_back(p->val);
                if (path_sum == sum && !p->left && !p->right){
                    matrix.push_back(path);
                }
                
                p = p->left;
            }
            if (!s.empty()){
                p = s.top();
                s.pop();
                
                if (p->left){
                    path_sum -= p->val;
                    path.pop_back();
                }
                
                p = p->right;
            }
        }
    }
    
    void pathSum2(TreeNode * root, int sum){
        if (!root){
            return;
        }
        
        path.push_back(root->val);
        
        if (!root->left && !root->right){
            if (sum == root->val){
                matrix.push_back(path);
            }
            return;
        }
         
        pathSum2(root->left,sum - root->val);
        
        path.pop_back();
        
        pathSum2(root->right,sum - root->val);
    }
    void pathSum4(TreeNode * root, int sum, int idx){
        if (!root){
            return;
        }
        
        if (path.size() != idx + 1){
            path.resize(idx + 1);
        }
        path[idx++] = root->val;
        if (!root->left && !root->right){
            if (sum == root->val){
                matrix.push_back(path);
            }
            return;
        }
        
        pathSum4(root->left, sum - root->val, idx);
        pathSum4(root->right, sum - root->val, idx);
        
    }
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        path.clear();
        matrix.clear();
        
        pathSum4(root,sum, 0);
        
        return matrix;
    }
};

PathSum2 and pathSum3 both are not right solution.because, we can not keep the path in right way...late I will fix them.

pathSum4 is good, since using idx as a little tricky