Binary Tree Path Sum II

public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
    public List<List<Integer>> binaryTreePathSum2(TreeNode root, int target) {
        /**思路
         * 起点不一定在root上面，那么可以从叶子节点逐步向上走
         * for循环找到等于target的路径输出*/
         //用result存最后结果，用buffer记录路径
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        ArrayList<Integer> buffer = new ArrayList<Integer>();
        if (root == null)
            return results;
            //这道题用的traverse，起点，终点，暂存，当前层数，结果
        findSum(root, target, buffer, 0, results);
        return results;
    }

    public void findSum(TreeNode head, int sum, ArrayList<Integer> buffer, int level, List<List<Integer>> results) {
        if (head == null) return;
        int tmp = sum;
        buffer.add(head.val);
        for (int i = level;i >= 0; i--) {
            tmp -= buffer.get(i);
            if (tmp == 0) {
                List<Integer> temp = new ArrayList<Integer>();
                for (int j = i; j <= level; ++j)
                    temp.add(buffer.get(j));
                results.add(temp);
            }
        }
        findSum(head.left, sum, buffer, level + 1, results);
        findSum(head.right, sum, buffer, level + 1, results);
        buffer.remove(buffer.size() - 1);
    }
}