題目：
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] ]
題解：
這道題除了要判斷是否有這樣的一個path sum，還需要把所有的都可能性結果都返回，所以就用傳統的DFS遞歸解決子問題。代碼如下：
1
public
void pathSumHelper(TreeNode root,
int sum, List <Integer> sumlist, List<List<Integer>> pathlist){
2 if(root== null)
3 return;
4 sumlist.add(root.val);
5 sum = sum-root.val;
6 if(root.left== null && root.right== null){
7 if(sum==0){
8 pathlist.add( new ArrayList<Integer>(sumlist));
9 }
10 } else{
11 if(root.left!= null)
12 pathSumHelper(root.left,sum,sumlist,pathlist);
13 if(root.right!= null)
14 pathSumHelper(root.right,sum,sumlist,pathlist);
15 }
16 sumlist.remove(sumlist.size()-1);
17 }
18
19 public List<List<Integer>> pathSum(TreeNode root, int sum) {
20 List<List<Integer>> pathlist= new ArrayList<List<Integer>>();
21 List<Integer> sumlist = new ArrayList<Integer>();
22 pathSumHelper(root,sum,sumlist,pathlist);
23 return pathlist;
24 }
2 if(root== null)
3 return;
4 sumlist.add(root.val);
5 sum = sum-root.val;
6 if(root.left== null && root.right== null){
7 if(sum==0){
8 pathlist.add( new ArrayList<Integer>(sumlist));
9 }
10 } else{
11 if(root.left!= null)
12 pathSumHelper(root.left,sum,sumlist,pathlist);
13 if(root.right!= null)
14 pathSumHelper(root.right,sum,sumlist,pathlist);
15 }
16 sumlist.remove(sumlist.size()-1);
17 }
18
19 public List<List<Integer>> pathSum(TreeNode root, int sum) {
20 List<List<Integer>> pathlist= new ArrayList<List<Integer>>();
21 List<Integer> sumlist = new ArrayList<Integer>();
22 pathSumHelper(root,sum,sumlist,pathlist);
23 return pathlist;
24 }