113. Path Sum II java solutions

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 a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        if(root == null) return ans;
        List<Integer> tmp = new ArrayList<Integer>();
        recursive(root,sum,tmp);
        return ans;
    }
    
    public void recursive(TreeNode root, int sum, List<Integer> tmp){
        if(root.val == sum && root.left == null && root.right == null){
            tmp.add(root.val);
            ans.add(tmp);
            return;
        }
        if(root.left != null){
            List<Integer> tmp1 = new ArrayList<Integer>(tmp);//需要使用新的list，不然原先的tmp修改会影响到right
            tmp1.add(root.val);
            recursive(root.left,sum-root.val,tmp1);
        }
        if(root.right != null){
            List<Integer> tmp2 = new ArrayList<Integer>(tmp);
            tmp2.add(root.val);
            recursive(root.right,sum-root.val,tmp2);
        }
    }
}