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
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        ArrayList<Integer> result= new ArrayList<Integer>();
        
        pSum(root, sum, results, result);
        return results;
    }

    public void pSum(TreeNode root, int sum, List<List<Integer>> results, ArrayList<Integer> result){
         if((root==null)&&(result.size()==0)){
                return;
       
        }else if((root.val==sum)&&(root.left==null&&root.right==null)){
            result.add(root.val);
            results.add(result);
            return;
        }else{
            if(root.left!=null&&root.right!=null){
                result.add(root.val);
                ArrayList<Integer> res= new ArrayList<Integer>();
                for(int i=0;i<result.size();i++){
                    res.add(result.get(i));
                }
                pSum(root.left, sum-root.val, results, result);
                pSum(root.right, sum-root.val, results, res);
                return ;
            }
            else if(root.left!=null&&root.right==null){
                result.add(root.val);
                pSum(root.left, sum-root.val, results, result);
                //pSum(root.right, sum-root.val, results, result);
                return ;
            }
            else if(root.left==null&&root.right!=null){
                result.add(root.val);
                //pSum(root.left, sum-root.val, results, result);
                pSum(root.right, sum-root.val, results, result);
                return;
            }
            return ;
            
        }
    }
}