int main() {
ifstream in("Stack1.cpp");
Stack1 textlines; // Try the different versions
// Read file and store lines in the stack:
string line;
while(getline(in, line))
textlines.push(line + "\n");
// Print lines from the stack and pop them:
while(!textlines.empty()) {
cout << textlines.top();
textlines.pop();
}
} ///:~
int main()
{
int InD[5010];
int PoD[5010];//将中序遍历序列和后续遍历序列放在InD和PoD中
int i, j , num;
BinNode<int>* p = new BinNode<int>, *root, *q;
Stack1 S1;
p->data = -1;
S1.push(p);
scanf("%d", &num);
for(i = 0; i < num; i++)
scanf("%d", &InD[i]);
for(i = 0; i < num; i++)
scanf("%d", &PoD[i]);
i = num - 1;
j = num - 1;
p = new BinNode<int>;
p->data = PoD[i];//后续序列的最后一个节点为根节点，确定root；
root = p; //p指向根节点
while(i >= 0 || j >= 0)
{
S1.push(p);//PoD[i]进栈
while(PoD[i] != InD[j])//而中序遍历序列中的最后一个节点为该二叉树最右下的节点InD[j]，在存放中序序列的数组InD中找到该节点，
{
i--;
p = new BinNode<int>;
p->data = PoD[i];
S1.top()->rChild = p; //PoD[i-1]的右孩子为PoD[i]
S1.push(p);//PoD[i]进栈
}
S1.top()->rChild = NULL; //PoD[i]的右孩子为空
i--;
j--;
q = S1.pop(); //退栈
while((j >= 0) && (InD[j] == S1.top()->data))
{
q->lChild = NULL; //q的左孩子为空
q = S1.pop();
j--;
}
if(i >= 0 || j >= 0)
{
p = new BinNode<int>;
p->data = PoD[i]; //q的左孩子为PoD[i]
q->lChild = p;
}
else q->lChild = NULL;
}
Stack1 S2; //辅助栈,利用中序遍历验证正确性
p = root;
i = 0;
while (true)
if (p)
{
S2.push(p); //根节点进栈
p = p->lChild; //深入遍历左子树
}
else if (!S2.empty())
{
p = S2.pop(); //尚未访问的最低祖先节点退栈
if(p->data != InD[i])
{
cout << -1;
return 0;
}
p = p->rChild; //遍历祖先的右子树
i++;
}
else break;
Stack1 S; //辅助栈
p = root;
i = 0;
if (p) S.push(p); //根节点入栈
while (!S.empty())
{
//在栈变空之前反复循环
p = S.pop();
if(i < num - 1)
{
printf("%d ", p->data); //弹出并访问当前节点，其非空孩子的入栈次序为
i++;
}
else
printf("%d", p->data);
if (HasRChild(*p))
S.push(p->rChild);
if (HasLChild(*p))
S.push(p->lChild); //先右后左
}
}
