3.9已知二叉树的 前序遍历和中序遍历的结果，重构出原二叉树

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#include<iostream>
using namespace std;
struct Node
{
	char value;
	Node *left;
	Node *right;
};
//重构函数核心
Node *Constructcore(char *start_preorder,char *end_preorder,char *start_inorder,char *end_inorder)
{
	char rootvalue=start_preorder[0];
	Node *root=new Node;
	root->value=rootvalue;
	root->left=root->right=NULL;
	if(start_preorder==end_preorder)
	{
		if(start_inorder==end_inorder&&*start_inorder==*start_preorder)
		{
			return root;
		}
		else
			throw std::exception("invalid inout");
	}
	//下面开始在中序遍历中找到根节点的值
	char *rootinorder=start_inorder;
	while(rootinorder<end_inorder&&*rootinorder!=rootvalue)
	{
		rootinorder++;
	}
	if(rootinorder==end_inorder&&*rootinorder!=rootvalue)
	{
		throw std::exception("invalid input");
	}
	int left_length=rootinorder-start_inorder;
	char *left_preorder_end=start_preorder+left_length;
	if(left_length>0)
	{
		root->left=Constructcore(start_preorder+1,left_preorder_end,start_inorder,rootinorder-1);
	}
	if(left_length<end_preorder-start_preorder)
	{
		root->right=Constructcore(left_preorder_end+1,end_preorder,rootinorder+1,end_inorder);
	}
	return root;
}
//重构函数
Node *construct(char *preorder,char *inorder,int length)
{
	if(preorder==NULL||inorder==NULL||length<=0)
		return NULL;
	return Constructcore(preorder,preorder+length-1,inorder,inorder+length-1);
}
void inorder_(Node *node)
{
	if(node==NULL)
		return ;
	cout<<node->value<<" ";
	inorder_(node->left);
	inorder_(node->right);
}
int main()
{
	char preorder[6]={'a','b','d','c','e','f'};
	char inorder[6]={'d','b','a','e','c','f'};
	Node *root=construct(preorder,inorder,6);
	inorder_(root);
	cout<<endl;
	system("pause");
	return 0;
}

3.9已知二叉树的 前序遍历和中序遍历的结果，重构出原二叉树

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原文地址：http://blog.csdn.net/qq_22335577/article/details/45332223