二叉树的先序中序后序遍历问题

已知先序和中序或中序和后序序列建立二叉树并遍历

#include<iostream>
#include<queue>
using namespace std;

typedef struct BTNode {
	char data;
	struct BTNode *lchild, *rchild;
}BTNode, *BTree;
queue<BTree> Q;

void DLR(BTree T)
{
	if (T)
	{
		cout << T->data<<" ";
		DLR(T->lchild);
		DLR(T->rchild);
	}
}

void LDR(BTree T)
{
	if (T)
	{
		LDR(T->lchild);
		cout << T->data << " ";
		LDR(T->rchild);
	}
}

void LRD(BTree T)
{
	if (T)
	{
		LRD(T->lchild);
		LRD(T->rchild);
		cout << T->data<<" ";
	}
}

void LevelOrderTraverse(BTree T)
{
	cout << T->data << " ";
	if(T->lchild)	Q.push(T->lchild);
	if(T->rchild)	Q.push(T->rchild);
	Q.pop();
	if (!Q.empty())LevelOrderTraverse(Q.front());
}

void Create_posOrder(BTree &T,char *pre,char *mid,int n)//已知先序、中序时建立二叉树
{
	T = new BTNode;
	if (n <= 0)
	{
		T = NULL;
		return;
	}
	T->data = *pre;

	char *pos;
	int i;
	for (pos = mid; pos < mid + n; pos++)
	{
		if (*pos == *pre) break;
	}
	i = pos - mid;
	Create_posOrder(T->lchild, pre + 1,mid,i);
	Create_posOrder(T->rchild, pre+i + 1,pos + 1, n - i - 1);
}

void Create_preOrder(BTree &T,char *mid,char *pos,int n)//已知中序、后序时建立二叉树
{
	T = new  BTNode;
	if (n <= 0)
	{
		T = NULL;
		return;
	}
	char *pre;
	int i;
	for (pre = mid; pre < mid + n; pre++)
	{
		if (*pre == *(pos + n - 1)) break;
	}
	i = pre - mid;
	T->data = *pre;
	Create_preOrder(T->lchild, mid ,pos,i);
	Create_preOrder(T->rchild, pre + 1,pos+i,n-i-1);
}

int main()
{
	char pre[] = "ABDEGCF";//先序序列
	char mid[] = "DBGEACF";//中序序列
	char pos[] = "DGEBFCA";//后序序列
	BTree T1,T2;
	
	Create_posOrder(T1, pre, mid, strlen(pre));
	cout << "先序遍历：";
	DLR(T1);
	cout << endl;
	cout << "中序遍历：";
	LDR(T1);
	cout << endl;
	cout << "后序遍历：";
	LRD(T1);
	cout << endl;
	cout << "层序遍历：";
	Q.push(T1);
	LevelOrderTraverse(Q.front());
	cout << endl;
	cout << endl;

	Create_preOrder(T2, mid, pos, strlen(pos));
	cout << "先序遍历：";
	DLR(T2);
	cout << endl;
	cout << "中序遍历：";
	LDR(T2);
	cout << endl;
	cout << "后序遍历：";
	LRD(T2);
	cout << endl;
	cout << "层序遍历：";
	Q.push(T2);
	LevelOrderTraverse(Q.front());
	cout << endl;
	cout << endl;

	return 0;
}