path-sum Java code

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 andsum = 22,

5

/ \

4 8

/ / \

11 13 4

/ \ / \

7 2 5 1

return

[

[5,4,11,2],

[5,8,4,5]

]

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
if(root == null)return ans;
if(root.left == null){
if(root.right == null){
if(root.val != sum)return ans;
ArrayList<Integer> list = new ArrayList<Integer>();
list.add(root.val);
ans.add(list);
return ans;
}else{
ArrayList<ArrayList<Integer>> right = pathSum(root.right, sum - root.val);
for(ArrayList<Integer> list : right){
list.add(0, root.val);
ans.add(list);
}
return ans;
}
}else{
if(root.right == null){
ArrayList<ArrayList<Integer>> left = pathSum(root.left, sum - root.val);
for(ArrayList<Integer> list : left){
list.add(0, root.val);
ans.add(list);
}
return ans;
}else{
ArrayList<ArrayList<Integer>> left = pathSum(root.left, sum - root.val);
for(ArrayList<Integer> list : left){
list.add(0, root.val);
ans.add(list);
}
ArrayList<ArrayList<Integer>> right = pathSum(root.right, sum - root.val);
for(ArrayList<Integer> list : right){
list.add(0, root.val);
ans.add(list);
}
return ans;

}
}
}
}