重建二叉树

根据输入的前序序列和中序序列，判断能否重建二叉树，能则输出后序遍历序列，否则输出“No”。

#include <iostream>
using namespace std;

int node_num = 0;
int pre_tree[1024] = {0};
int in_tree[1024] = {0};
int post_tree[1024] = {0};

bool Rebulid(int pre_start,int pre_end,int in_start,int in_end,int post_start,int post_end)
{
	if(pre_end - pre_start != in_end - in_start)    //左右子树数量不等
	{
		return false;
	}

	int root = pre_tree[pre_start];   //根节点
	post_tree[post_end] = root;		  //后序遍历根节点在最后一个

	int i = 0;
	for(i = in_start; i <= in_end; ++i)  //寻找中序遍历根节点位置
	{
		if(in_tree[i] == root)
		{
			break;
		}
	}

	if(i > in_end)   //没有找到返回失败
	{
		return false;
	}

	int len1 = i - in_start;    //左子树长度
	int len2 = in_end - i;      //右子树长度

	bool left_flag = false;     //左子树能否重建标记
	bool right_flag = false;	//右子树能否重建标记

	if(len1 == 0)               //所有左子树重建完成
	{
		left_flag = true;
	}
	else
	{
		//递归判断左子树能否重建
		left_flag = Rebulid(pre_start+1,pre_start+len1,in_start,in_start+len1-1,post_start,post_start+len1-1);
	}

	if(len2 == 0)   //所有右子树重建完成
	{
		right_flag = true;
	}
	else
	{
		//递归判断右子树能否重建
		right_flag = Rebulid(pre_start+len1+1,pre_end,in_start+len1+1,in_end,post_start+len1,post_end-1);
	}

	return (left_flag && right_flag);  //左右子树都能重建能返回true

}

int main()
{
	int i = 0;
	while (cin>>node_num)
	{
		for(i = 0; i < node_num; ++i)
		{
			cin >>pre_tree[i];
		}

		for(i = 0; i < node_num; ++i)
		{
			cin >>in_tree[i];
		}


		if(Rebulid(0,node_num-1,0,node_num-1,0,node_num-1))
		{
			for(i = 0; i < node_num; ++i)
			{
				cout<<post_tree[i]<<" ";
			}
			cout<<endl;
		}
		else
		{
			cout<<"No"<<endl;
		}
	}

	return 0;
}