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]
] 

recursion解法，樹的解法比較固定，分為base case和recursion rule，然後定好引數，在需要的時候呼叫自身即可。

public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        helper(res, root, new
 
 ArrayList<Integer>(), sum);
        return res;
    }
    public void helper(List<List<Integer>> res, TreeNode root, ArrayList<Integer> path, int sum) {
        //base case: leaf
        if (root.left == null && root.right == null) {
            if (root.val == sum) {
                path.
 
add(root.val);
                res.add(new ArrayList<Integer>(path));
                path.remove(path.size() - 1);
            }
            return;
        }
        //recursion rule: non-leaf
        path.add(root.val);
        if (root.left != null) helper(res, root.left, path, sum - root.val);
        if (root.right != null) helper(res, root.right, path, sum - root.val);
        path.remove(path.size() - 1);
        return;
    }