[Leetcode] Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

Analysis:

The idea is very simple: Just Preorder visit the tree and for each node, treat it as root and count the number of path which is equal to sum. There should be a better way to remove the duplicate visit?  

    int helper(TreeNode root, int sum, int total) {
        if (root == null)
            return 0;
        else {
            total += root.val;
            int count = sum == total ? 1 : 0;
            count += helper(root.left, sum, total);
            count += helper(root.right, sum, total);
            return count;
        }
    }
    
    public int pathSum(TreeNode root, int sum) {
        if (root == null)
            return 0;
        else {
            int count = helper(root, sum, 0);
            return count + pathSum(root.left, sum) + pathSum(root.right, sum);
        }
    }