【算法】非递归的方式生成树

【算法】非递归的方式生成树

 一种不同方式来生成树，感觉蛮有意思的，在此记录

 public class TreeServices
    {
        /// <summary>
        /// 以非递归的方式生成树
        /// </summary>
        /// <param name="table"></param>
        /// <returns></returns>
        public static List<TreeNode> ToTree(DataTable table)
        {
            var list = new List<TreeNode>();
            if (table.IsEmpty()) return list;
            var dic = TreeNode.FillModel(table);
            foreach (var pair in dic)
            {
                var pid = pair.Value.ParentId;
                if (pid == 0)
                {
                    list.Add(pair.Value);
                    continue;
                }
                if (dic.ContainsKey(pid))
                {
                    dic[pid].Children.Add(pair.Value);
                }
            }
            return list;
        }
    }

    public class TreeNode
    {
        public TreeNode()
        {
            Children = new List<TreeNode>();
        }

        /// <summary>
        /// 自增Id
        /// </summary>
        public int NodeId { get; set; }

        /// <summary>
        /// 关联Id，形式1-2-3-4（该节点上一节点为4，再上一级为3，...）
        /// </summary>
        public string RelationId { get; set; }

        /// <summary>
        /// 上一个节点的Id
        /// </summary>
        public int ParentId
        {
            get
            {
                if (RelationId.IsEmpty()) return 0;
                var s = RelationId.Split('-');
                return s[s.Length - 1].ToInt32();
            }
        }

        /// <summary>
        /// 是否有下一节点
        /// </summary>
        public bool HasChild
        {
            get { return Children.IsEmpty(); }
        }

        /// <summary>
        /// 子节点集合
        /// </summary>
        public List<TreeNode> Children { get; set; }

        /// <summary>
        /// DataTable To Dictionary
        /// </summary>
        /// <param name="table"></param>
        /// <returns></returns>
        public static Dictionary<int, TreeNode> FillModel(DataTable table)
        {
            var list = new Dictionary<int, TreeNode>();
            if (table == null || table.Rows.Count == 0) return list;
            foreach (DataRow row in table.Rows)
            {
                var model = new TreeNode
                {
                    NodeId = row["Id"].ToInt32(),
                    RelationId = row["RelationId"].ToString()
                };
                list.Add(model.NodeId, model);
            }
            return list;
        }
    }