# 03-树3 Tree Traversals Again

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。

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

### 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 } 

 
posted on 2016-03-20 00:30  kuotian  阅读(265)  评论(0编辑  收藏  举报