113. 路徑總和 II

這題跟上個題的區別112. 路徑總和,需要儲存下路徑，且有可能出現多條路徑。

在前一個題的基礎上加上回溯即可

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> ansList = new ArrayList<>();
        if (root == null) return ansList;
        List<Integer> curList = new ArrayList<>();

        findPath(root, sum, curList, ansList);

        return ansList;
    }

    private void findPath(TreeNode root, int sum, List<Integer> curList, List<List<Integer>> ansList) {
        if (root == null) return;
        curList.add(root.val);
        if (root.left == null && root.right == null && sum == root.val) {
            List<Integer> tmp = new ArrayList<>(curList);
            ansList.add(tmp);
        } else {
            findPath(root.left, sum - root.val, curList, ansList);
            findPath(root.right, sum - root.val, curList, ansList);
        }
        curList.remove(curList.size() - 1);
    }
}