crack the code intervie 4.8

You are given a binary tree in which each node contains a value. Design an algorithm to print all paths which sum up to that value. Note that it can be any path in the tree - it does not have to start at the root.

#include <list>

#include <iostream>
using namespace std;


struct TreeNode
{
    int value;
    TreeNode * left;
    TreeNode * right;
    TreeNode(int v)
    {
        value = v;
        left = NULL;
        right = NULL;
    }
    TreeNode(int v, TreeNode * l, TreeNode * r)
    {
        value = v;
        left = l;
        right = r;
    }
};


void mprint(list<int> path, int length);


void findSum(TreeNode * node, int sum_value, list<int> path, int length)
{
    if (node == NULL)
        return;
    path.push_front(node->value);
    int tmp = sum_value;
    list<int>::iterator iter;
    int i = 0;
    for (iter = path.begin(); iter != path.end(); ++iter)
    {
        tmp -= *iter;
        if (tmp == 0)
            mprint(path, i);
        i ++;
    }
    list<int> c1(path);
    list<int> c2(path);
    findSum(node->left, sum_value, c1, length + 1);
    findSum(node->right, sum_value, c2, length + 1);
    
}


void mprint(list<int> path, int length)
{
    list<int>::iterator iter;
    for(iter = path.begin(); length >= 0; length --)
    {
        cout<<*iter<<" ";
        iter ++;
    }
    cout<<endl;
}


int main()
{
    TreeNode * node0 = new TreeNode(2);
    TreeNode * node1 = new TreeNode(3, node0, NULL);
    TreeNode * node2 = new TreeNode(-4, node1, NULL);
    TreeNode * node3 = new TreeNode(3, node2, NULL);
    TreeNode * node4 = new TreeNode(1, node3, NULL);
    TreeNode * node5 = new TreeNode(2, node4, NULL);
    list<int> path;
    findSum(node5, 5, path, 0);
    return 0;
}