二叉樹後續遍歷的非遞歸循環C# 實現

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<span style="font-size: 18px; font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);"></span><pre code_snippet_id="495329" snippet_file_name="blog_20141024_1_8664932" name="code" class="csharp">對於二叉樹的遍歷循環一直覺得是自己的軟肋， 在網上看到許多一些交叉， 總覺得只是囫圇吞棗， 強行記住， 卻隱隱覺得自己沒有真正理解其中的真諦

於是特地抽出時間對二叉樹的遍歷用代碼實現了一遍， 找找手感。 二叉樹的前序遍歷， 和中序遍歷整體感覺上都挺好實現的， 後續遍歷由於需要兩次返回節點， 頗有挑戰性。 

首先溫習一下後續遍歷的規則

        public void Recurrence_travel(Bittree ptree)
        {
            if (ptree == null)
            {
                return;
            }
            Recurrence_travel(ptree.plefttree);
            Recurrence_travel(ptree.prighttree);
            Console.WriteLine(ptree.keyvalue.ToString());
        }

二叉樹的循環實現：

        public void Loop_travel(Bittree ptree)
        {
            if (ptree == null)
            {
                return;
            }
            Poststack treelist = new Poststack();
            int counts=-1;

            while (ptree != null || treelist.Get_size() != 0)
            {
                while (ptree != null)
                {
                    if (ptree.plefttree == null && ptree.prighttree == null)
                    {
                        Console.WriteLine(ptree.keyvalue.ToString());
                        break;                      // jump out  first loop 
                    }
                    if (ptree.plefttree != null)
                    {
                        treelist.Push(ptree);
                        ptree = ptree.plefttree;
                        continue;                   //ship into next loop
                    }
                    if (ptree.plefttree==null && ptree.prighttree != null)
                    {
                        treelist.Push(ptree);
                        break;
                    }

                }
                ptree = treelist.Pop(ref counts);
                if (counts == 0)                  // first time to get back
                {
                    ptree = ptree.prighttree;      // turn right to travel 
                    continue;
                }

                if (counts == 1)                     // second time to get back 
                {
                    Console.WriteLine(ptree.keyvalue.ToString());
                    ptree = null; // ensure it would not enter second loop 
                }

            }

        }

輔助堆棧：

    class Poststack
    {
        // second paramter could be counts for the pop out 
        private Dictionary<Bittree, int> treedict=new Dictionary<Bittree, int>();
        public Poststack()
        {

        }

        public void Push(Bittree ptree)
        {
            if (ptree == null)
            {
                return;
            }

            treedict.Add(ptree, 0);
        }


        public Bittree  Pop(ref int pcounts)
        {
            if (treedict.Count() ==0)
            {
                return null;
            }
            Bittree tptree = treedict.Last().Key;
            pcounts = treedict.Last().Value;
            if (pcounts == 0)
            {
                treedict.Remove(tptree);
                treedict.Add(tptree, pcounts + 1);
                return tptree;
            }

            if (pcounts == 1)    // second travel 
            {
                treedict.Remove(tptree);
                return tptree;
            }

            return null;
        }

        public int Get_size()
        {
            return treedict.Count();
        }



    }