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]
]

tag: binary tree, DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) return ans;

        dfs(root, new ArrayList<>(), ans, 0, sum);

        return ans;
    }

    private void dfs(TreeNode root, List<Integer> temp, List<List<Integer>> ans, int path, int sum){
        if (root == null) return;

        temp.add(root.val);
        path += root.val;
        if (root.left == null && root.right == null){
            //System.out.println("path: " + path + " temp: " + temp.toString());
            if (path == sum){
                ans.add(new ArrayList<>(temp));
            }
        }

        dfs(root.left, temp, ans, path, sum);
        dfs(root.right, temp, ans, path, sum);
        temp.remove(temp.size() - 1);
        path -= root.val;
    }
}