假設已經有了前序走訪和中序遍曆的結果，通過一個演算法重建這棵樹

分析與解法

前序: a b c d e f

後序: d b a e c f

“a”是前序走訪節點的第一個元素,它把中序遍曆的結果分為“db”和“ecf”兩個部分，這兩部分也是“a”的左右子樹的遍曆結果。

如果能夠找到前序走訪中對應的左子樹和右子樹，就可以把“a”作為當前的根節點，然後依次遞迴下去，這樣就能夠依次恢複左子樹和右子樹的遍曆結果。

using System;using System.Collections.Generic;using System.Text;namespace ConsoleApp{    class Program    {        static void Main(string[] args)        {            TreeNode node1 = new TreeNode();            node1.Data = 1;            TreeNode node2 = new TreeNode();            node2.Data = 2;            TreeNode node3 = new TreeNode();            node3.Data = 3;            TreeNode node4 = new TreeNode();            node4.Data = 4;            TreeNode node5 = new TreeNode();            node5.Data = 5;            TreeNode node6 = new TreeNode();            node6.Data = 6;            TreeNode node7 = new TreeNode();            node7.Data = 7;            TreeNode node8 = new TreeNode();            node8.Data = 8;            TreeNode node9 = new TreeNode();            node9.Data = 9;            TreeNode[] nodes1 ={ node1, node2, node4, node7, node3, node5, node8, node6, node9 };            TreeNode[] nodes2 ={ node7, node4, node2, node1, node8, node5, node3, node6, node9 };            TreeNode root = ReBuild.ReBuildTree(nodes1, nodes2, 0, 0, nodes1.Length);            Tree tree2 = new Tree();            tree2.Root = root;            Tree.Print2(tree2);            Console.ReadKey();        }    }    class Tree    {        public TreeNode Root;        ///         /// 廣度優先遍曆二叉樹        ///         public static void Print2(Tree tree)        {            if (tree.Root == null)            {                return;            }            else            {                Queue> queue = new Queue>();                queue.Enqueue(tree.Root);                while (queue.Count != 0)                {                    int currentLevelCount = queue.Count;                    for (int i = 0; i < currentLevelCount; i++)                    {                        TreeNode node = queue.Dequeue();                        if (node.Data != null)                        {                            Console.Write(node.Data);                        }                        if (node.LeftSon != null)                        {                            queue.Enqueue(node.LeftSon);                        }                        if (node.RightSon != null)                        {                            queue.Enqueue(node.RightSon);                        }                    }                    Console.WriteLine();                }            }        }    }    class TreeNode    {        public T Data;        public TreeNode LeftSon;        public TreeNode RightSon;    }    class ReBuild    {        public static TreeNode ReBuildTree(TreeNode[] preOrder, TreeNode[] inOrder, int preIndex, int inIndex, int length)        {            if (preOrder == null || length == 0)            {                return null;            }            else if (preOrder.Length != inOrder.Length)            {                return null;            }            else if (length == 1)            {                return preOrder[preIndex];            }            for (int i = inIndex; i < inIndex + length; i++)            {                if (preOrder[preIndex] == inOrder[i])                {                    preOrder[preIndex].LeftSon = ReBuildTree(preOrder, inOrder, preIndex + 1, inIndex, i - inIndex);                    preOrder[preIndex].RightSon = ReBuildTree(preOrder, inOrder, preIndex + i - inIndex + 1, i + 1, length - i + inIndex - 1);                }            }            return preOrder[preIndex];        }    }}