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Path Sum II leetcode java
题目:
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]]
题解:
这道题除了要判断是否有这样的一个path sum,还需要把所有的都可能性结果都返回,所以就用传统的DFS递归解决子问题。代码如下:
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 }
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