LintCode 376 Binary Tree Path sum

LintCode 376 Binary Tree Path sum

Example

Given a binary tree, and target = 5:

1
/ \

2 4

/ \

2 3

return

[

[1, 2, 2],

[1, 4]

]

/**
* Definition of TreeNode:
* class TreeNode {
* public:
*     int val;
*     TreeNode *left, *right;
*     TreeNode(int val) {
*         this->val = val;
*         this->left = this->right = NULL;
*     }
* }
*/
class Solution {
public:
/**
* @param root the root of binary tree
* @param target an integer
* @return all valid paths
*/
void Travel(TreeNode* node, vector<int>& path, int sum, vector<vector<int>>& result, int target)
{
if (sum == target && node->left == NULL&& node->right == NULL)
{
result.push_back(path);
}
if (node->left != NULL)
{
sum += node->left->val;
path.push_back(node->left->val);
Travel(node->left, path, sum, result, target);
path.pop_back();
sum -= node->left->val;
}
if (node->right != NULL)
{
sum += node->right->val;
path.push_back(node->right->val);
Travel(node->right, path, sum, result, target);
path.pop_back();
sum -= node->right->val;
}
}

vector<vector<int>> binaryTreePathSum(TreeNode *root, int target)
{
// Write your code here

vector<vector<int>>* result = new vector<vector<int>>;
if (root == NULL)
return *result;
vector<int>* path = new vector<int>;
path->push_back(root->val);
int sum = root->val;
Travel(root, *path, sum, *result, target);
return *result;
}
};