题解 | #二叉树中和为某一值的路径(二)#（阿飞算法）

JZ34 二叉树中和为某一值的路径(二)

方法1:回溯-递减

        ArrayList<ArrayList<Integer>> res = new ArrayList<>();

        public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int targetSum) {
            dfs(root, targetSum, new ArrayList<>());
            return res;
        }

        private void dfs(TreeNode root, int sum, ArrayList<Integer> sub) {
            if (root == null) {
                return;
            }
            sub.add(root.val);
            if (root.left == null && root.right == null && sum == root.val) res.add(new ArrayList<>(sub));
            dfs(root.left, sum - root.val, sub);
            dfs(root.right, sum - root.val, sub);
            sub.remove(sub.size() - 1);
        }

方法2:回溯-累加

   ArrayList<ArrayList<Integer>>  res = new ArrayList<>();

        public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int sum) {
            if (root == null) return res;
            dfs(root,sum,0,new ArrayList<>());
            return res;
        }

        private void dfs(TreeNode root, int sum, int total, ArrayList<Integer> sub) {
            if (root == null) return;
            sub.add(root.val);
            total += root.val;
            if (root.left == null && root.right == null) {
                if (sum == total) {
                    res.add(new ArrayList<>(sub));
                }
                //这一步是为了移除这个叶子节点的值，不管这个叶子节点是否满足条件
                sub.remove(sub.size() - 1);
                return;
            }
            dfs(root.left, sum, total, sub);
            dfs(root.right, sum, total, sub);
            sub.remove(sub.size() - 1);
        }