pta 编程题8 Tree Traversals Again

其它pta数据结构编程题请参见：pta

这次的作业考察的是树的遍历。

题目的输入通过栈的pop给出了树的中序遍历的顺序。根据push和pop的顺序构造树的方法为：定义一个变量father来确定父节点，如果父节点还没有pop，那么push操作就构造父节点的左子树，否则构造父节点的右子树；定义一个栈用来确定pop操作弹出的节点，将father赋值为pop的节点，并将此节点的flag值（用来标记已经pop过）标为1。

 #include <iostream>
 #include <string>
 #include <vector>
 using namespace std;

 struct Stack
 {
     int data[];
     int top = -;
 };

 struct Node
 {
     int left;
     int right;
     int flag;//has poped
 }tree[];

 Node initNode();
 void push(Stack &s, int i);
 int pop(Stack &s);
 void postOrder(int i, vector<int> &v);

 int main()
 {
     int num, t, i;
     int root, father = -;
     cin >> num;
     string s;
     Stack stack;
     for (i = ; i <  * num; i++)
     {
         cin >> s;
         if (s == "Push")
         {
             cin >> t;
             push(stack, t);
             if (i == ) root = t;
             else if (tree[father].flag == )
                 tree[father].left = t;
             else
                 tree[father].right = t;//父节点已经pop过只能向右节点push
             tree[t] = initNode();
             father = t;
         }
         else
         {
             father = pop(stack);
             tree[father].flag = ;
         }
     }
     vector<int> v;
     postOrder(root, v);
     for (int i = ; i < v.size(); i++)
     {
         if (i) cout << " ";
         cout << v[i];
     }
     return ;
 }

 Node initNode()
 {
     Node node;
     node.left = -;
     node.right = -;
     node.flag = ;
     return node;
 }

 void push(Stack &s, int i)
 {
     s.data[++s.top] = i;
 }

 int pop(Stack &s)
 {
     return s.data[s.top--];
 }

 void postOrder(int i, vector<int> &v)
 {
     if (tree[i].left != -) postOrder(tree[i].left, v);
     if (tree[i].right != -) postOrder(tree[i].right, v);
     v.push_back(i);
 }