Path Sum II 解答

 1 /**
 2  * Definition for a binary tree node.
 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     public List<List<Integer>> pathSum(TreeNode root, int sum) {
12         List<List<Integer>> result = new ArrayList<List<Integer>>();
13         List<Integer> prevList = new ArrayList<Integer>();
14         dfs(root, sum, result, prevList);
15         return result;   
16     }
17     
18     private void dfs(TreeNode root, int target, List<List<Integer>> result, List<Integer> prevList) {
19         if (root == null)
20             return;
21         prevList.add(root.val);
22         
23         if (root.left == null && root.right == null) {
24             if (root.val == target)
25                 result.add(new ArrayList<Integer>(prevList));
26         } else {
27             List<Integer> tmpList2 = new ArrayList<Integer>(prevList);
28             if (root.left != null)
29                 dfs(root.left, target - root.val, result, prevList);
30             if (root.right != null)
31                 dfs(root.right, target - root.val, result, tmpList2);
32         }
33         
34     }
35 }