03-树2. Tree Traversals Again (25)

03-树2. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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

 

    这题要做的是树的中缀转后缀，一开始思路有问题，没想明白，只过了两组数据，后来静下来好好想发现也不难。

题目中给出了中缀遍历树的过程，根据题目给出的过程用栈模拟建立树，然后后缀输出。

 1 #include <iostream>
 2 #include <stack>
 3  using  namespace std;
 4 
 5  struct node  // 树节点
 6  {
 7         int left;
 8         int right;
 9        node () // 结点初始化左右儿子为-1，表示左右结点都不存在
10         {
11             left=- 1;
12             right=- 1;
13        }
14 };
15 node no[ 50];
16 
17 stack< int> s; // 用栈模拟题目中给出的建树过程，栈顶元素为正在处理的结点，很方便的把每个结点的左右儿子构造好
18 
19  int t;
20  void print( int first) // 后缀输出结点
21  {
22      if (first==- 1)
23      return;
24     print(no[first].left);
25     print(no[first].right);
26      if (t== 0)
27     {
28              t= 1;
29     cout <<first;
30 }
31      else
32     cout<< "   "<<first;
33 }
34  int main()
35 {
36      int m,n,now,first; // now代表当前处理的结点，first代表树根
37       string str;
38      while (cin>>n)
39     {
40           t= 0;
41            while (!s.empty()) // 第一个肯定是树根
42            s.pop(); // 入栈
43            
44           cin>>str>>m;
45           first=m;
46           s.push(m);
47           now=m;
48            for ( int i= 1;i< 2*n;i++)
49           {
50                 
51                 cin>>str;
52                  if (str== " Push ")
53                 {
54                      cin>>m;
55                       if (no[now].left==- 1)
56                      no[now].left=m;
57                       else
58                      no[now].right=m;
59                      now=m; // 下一个要处理的结点为m
60                       s.push(m);
61                 }
62                  else
63                 {
64                     now=s.top(); // 下一个要处理的结点为s.top
65                      s.pop();
66                 }
67           }
68           print(first);
69           cout <<endl;
70     }
71      return  0;
72 }

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