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]] 1。下面的搜索总超时，可能我写的不好

1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     ArrayList
      
       
        > list=new ArrayList
        
         
          >();12      ArrayList
          
            arry=new ArrayList
           
            ();13 14 public List
            
             
              > pathSum(TreeNode root, int sum) {15 16 if(root==null) return (List)list;17 if(root.right==null&&root.left==null)18 {19 if(sum==root.val)20 {21 22 list.add(new ArrayList(arry));23 }24 }25 26 arry.add(root.val);27 pathSum(root.right,sum-root.val);28 29 30 pathSum(root.left,sum-root.val);31 32 33 return (List)list;34 35 36 37 38 39 }40 41 42 43 }