二进制重建

    1.先序遍历方式为：根->左->右.故1为根节点。

中序方式为：左->根->右,所以4,7,2为左子树上的结点，5,3,8,6为右子树的结点。

  2.经过步骤1，将问题变成两个子问题。我们先考虑其左子树。先序遍历结果为{2,4,7},中序遍历结果为{4,7,2}.故左子树的根结点为2，那么依据中序遍历结果，{4,7}为其左子树。

 3.经过步骤2，子问题又被分解为两个子问题。依据中序遍历结果，我们发现。{4,7}为其左子树，没有右子树，所以我们仅需考虑左子树。继续依照上面的方式：先序遍历结果为{4,7},说明根结点为4,中序遍历结果为{4,7}。说明7是右子树。

4.自此，左子树所有分析完，以下分析右子树，先序遍历结果为{3,5,6，8}，中序结果为{5,3,8,6}。说明3为根结点。5为左子树。{8,6}为右子树。再递推。6应该是根结点，8为左子树。终于得到二叉树：

/***************************************
重建二叉树
by Rowandjj
2014/7/19
***************************************/
#include<iostream>
#include<stdlib.h>
using namespace std;
typedef struct _BINNODE_//二叉树结点定义
{
	int data;
	struct _BINNODE_ *left;
	struct _BINNODE_ *right;
}BTNode,*BTree;
int CanReBuild;//1可重建，0 不可重建
//重建二叉树
void RebuildBTree(BTree *ppTree,int *pre,int *inv,int len)
{//pre为先序遍历数组，inv为中序遍历数组。len为数组长度
	
	if(pre == NULL || inv == NULL)
	{
		CanReBuild = false;
		return;
	}
	if(len <= 0)
	{
		return;
	}
	int i;
	for(i = 0; i < len; i++)
	{
		if(pre[0] == inv[i])
		{
			break;
		}
	}
	if(i >= len)
	{
		CanReBuild = false;
		return;
	}
	*ppTree = (BTNode*)malloc(sizeof(BTNode));
	if(*ppTree == NULL)
	{
		exit(-1);
	}
	(*ppTree)->data = pre[0];
	(*ppTree)->left = NULL;
	(*ppTree)->right = NULL;
	RebuildBTree(&(*ppTree)->left,pre+1,inv,i);
	RebuildBTree(&(*ppTree)->right,pre+i+1,inv+i+1,len-i-1);
}
//后序遍历
void BehindTraverse(BTree bTree)
{
	if(!bTree)
	{
		return;
	}
	if(bTree->left)
	{
		BehindTraverse(bTree->left);
	}
	if(bTree->right)
	{
		BehindTraverse(bTree->right);
	}
	cout<<bTree->data<<" ";
}
//销毁二叉树
void DestroyTree(BTree bTree)
{
	if(!bTree)
	{
		return;
	}
	if(bTree->left)
	{
		DestroyTree(bTree->left);
	}
	if(bTree->right)
	{
		DestroyTree(bTree->right);
	}
	free(bTree);
	bTree = NULL;
}
int main()
{
	BTree bTree = NULL;
	int m,i;
	int *pre,*inv;
	while(cin>>m)
	{
		pre = (int *)malloc(sizeof(int)*m);
		inv = (int *)malloc(sizeof(int)*m);
		if(!pre || !inv)
		{
			exit(-1);
		}
		for(i = 0; i < m; i++)
		{
			cin>>pre[i];
		}
		for(i = 0; i < m; i++)
		{
			cin>>inv[i];
		}
		CanReBuild = 1;
		RebuildBTree(&bTree,pre,inv,m);
		if(CanReBuild)
		{
			BehindTraverse(bTree);
			cout<<endl;
			DestroyTree(bTree);
		}
		else
		{
			cout<<"No"<<endl;
		}
		free(pre);
		free(inv);
		pre = NULL;
		inv = NULL;
	}
	return 0;
}