第四十三题 通过前序遍历和中序遍历构建二叉树

题目：通过树的前序遍历和中序遍历来确定树的结构并输出，通过递归的方法

#include <iostream>
//#include <boost\shared_ptr.hpp>
using namespace std;
struct NodeTree
{
	int m_value;
	NodeTree *m_left;
	NodeTree *m_right;
};
NodeTree* maketree(int *pre,int *mid,int length)
{
	if (pre==nullptr||mid==nullptr||length<=0)
	{
		cout<<"input is error"<<endl;
		return nullptr;
	}
	NodeTree *root=new NodeTree();
	//boost::shared_ptr<NodeTree> root(new NodeTree());
	root->m_value=*pre;
	root->m_left=nullptr;
	root->m_right=nullptr;
	if (length==1)
	{
		if (*pre==*mid)
		{
			return root;
		}
		else
		{
			cout<<"error"<<endl;
			return nullptr;
		}
	}
	int *midbeg=mid;
	while (*midbeg!=root->m_value&&midbeg<mid+length)
	{
		++midbeg;
	}
	if (midbeg==mid+length)
	{
		cout<<"error root"<<endl;
		return nullptr;
	}
	int rootlength=midbeg-mid;
	//构建左子树
	if (rootlength>0)
	{
		root->m_left=maketree(pre+1,mid,rootlength);
	}
	//构建右子树
	if (rootlength<length-1)
	{
		root->m_right=maketree(pre+rootlength+1,midbeg+1,length-1-rootlength);
	}
	return root;
}
void PrintTreeNode(NodeTree* pNode)
{
	if(pNode != nullptr)
	{
		printf("value of this node is: %d\n", pNode->m_value);

		if(pNode->m_left != nullptr)
			printf("value of its left child is: %d.\n", pNode->m_left->m_value);
		else
			printf("left child is null.\n");

		if(pNode->m_right != nullptr)
			printf("value of its right child is: %d.\n", pNode->m_right->m_value);
		else
			printf("right child is null.\n");
	}
	else
	{
		printf("this node is null.\n");
	}

	printf("\n");
}
void PrintTree(NodeTree* pRoot)
{
	PrintTreeNode(pRoot);

	if(pRoot != nullptr)
	{
		if(pRoot->m_left != nullptr)
			PrintTree(pRoot->m_left);

		if(pRoot->m_right != nullptr)
			PrintTree(pRoot->m_right);
	}
}
int main()
{
	const int length = 5;
	int preorder[length] = {1, 2, 3, 4, 5};
	int inorder[length] = {5, 4, 3, 2, 1};
	NodeTree *root=nullptr;
	root=maketree(preorder,inorder,length);
	PrintTree(root);
	delete root;
	root=nullptr;
	return 0;
}

解决