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。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 using namespace std; 5 6 struct Stack 7 { 8 int data[31]; 9 int top = -1; 10 }; 11 12 struct Node 13 { 14 int left; 15 int right; 16 int flag;//has poped 17 }tree[31]; 18 19 Node initNode(); 20 void push(Stack &s, int i); 21 int pop(Stack &s); 22 void postOrder(int i, vector<int> &v); 23 24 int main() 25 { 26 int num, t, i; 27 int root, father = -1; 28 cin >> num; 29 string s; 30 Stack stack; 31 for (i = 0; i < 2 * num; i++) 32 { 33 cin >> s; 34 if (s == "Push") 35 { 36 cin >> t; 37 push(stack, t); 38 if (i == 0) root = t; 39 else if (tree[father].flag == 0) 40 tree[father].left = t; 41 else 42 tree[father].right = t;//父节点已经pop过只能向右节点push 43 tree[t] = initNode(); 44 father = t; 45 } 46 else 47 { 48 father = pop(stack); 49 tree[father].flag = 1; 50 } 51 } 52 vector<int> v; 53 postOrder(root, v); 54 for (int i = 0; i < v.size(); i++) 55 { 56 if (i) cout << " "; 57 cout << v[i]; 58 } 59 return 0; 60 } 61 62 Node initNode() 63 { 64 Node node; 65 node.left = -1; 66 node.right = -1; 67 node.flag = 0; 68 return node; 69 } 70 71 void push(Stack &s, int i) 72 { 73 s.data[++s.top] = i; 74 } 75 76 int pop(Stack &s) 77 { 78 return s.data[s.top--]; 79 } 80 81 void postOrder(int i, vector<int> &v) 82 { 83 if (tree[i].left != -1) postOrder(tree[i].left, v); 84 if (tree[i].right != -1) postOrder(tree[i].right, v); 85 v.push_back(i); 86 }