push为前序遍历序列，pop为中序遍历序列。将题目转化为已知前序、中序，求后序。
前序GLR 中序LGR
前序第一个为G，在中序中找到G，左边为左子树L，右边为右子树R。
将左右子树看成新的树，同理。
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Figure 1
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N(≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
1 #include <iostream>
2 #include <cstdio>
3 #include <stack>
4 #include <string>
5 using namespace std;
6
7 #define MaxSize 30
8
9 #define OK 1
10 #define ERROR 0
11
12 int preOrder[MaxSize];
13 int inOrder[MaxSize];
14 int postOrder[MaxSize];
15
16 void postorderTraversal(int preNo, int inNo, int postNo, int N);
17
18 int main()
19 {
20 stack<int> stack;
21 int N; //树的结点数
22 cin >> N;
23 string str;
24 int data;
25 int preNo = 0, inNo = 0, postNo = 0;
26 for(int i = 0; i < N * 2; i++) { //push + pop = N*2
27 cin >> str;
28 if(str == "Push") { //push为前序序列
29 cin >> data;
30 preOrder[preNo++] = data;
31 stack.push(data);
32 }else{ //pop出的是中序序列
33 inOrder[inNo++] = stack.top();
34 stack.pop(); //pop() 移除栈顶元素（不会返回栈顶元素的值）
35 }
36 }
37 postorderTraversal(0, 0, 0, N);
38 for(int i = 0; i < N; i++) { //输出后序遍历序列
39 if(i == 0) //控制输出格式
40 printf("%d",postOrder[i]);
41 else
42 printf(" %d",postOrder[i]);
43 }
44 printf("\n");
45 return 0;
46 }
47
48 void postorderTraversal(int preNo, int inNo, int postNo, int N)
49 {
50 if(N == 0)
51 return;
52 if(N == 1) {
53 postOrder[postNo] = preOrder[preNo];
54 return;
55 }
56 int L, R;
57 int root = preOrder[preNo]; //先序遍历GLR第一个为根
58 postOrder[postNo + N -1] = root; //后序遍历LRG最后一个为根
59 for(int i = 0; i < N; i++) {
60 if(inOrder[inNo + i] == root) { //找到中序的根 左边为左子树 右边为右子树
61 L = i; //左子树的结点数
62 break;
63 }
64 }
65 R = N - L - 1; //右子树的结点数
66 postorderTraversal(preNo + 1, inNo, postNo, L); //同理，将左子树看成新的树
67 postorderTraversal(preNo + L + 1, inNo + L + 1, postNo + L, R);//同理，右子树
68 }