【树】7-5Tree Traversals Again

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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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

思路：

  • 通过观察可知入栈顺序即先序遍历，出栈顺序即中序遍历，所以在得知先序和中序遍历的基础上还原树，然后对该树进行后序遍历。
  • 其中，为了得到先序（中序）结果，对每一条命令进行判断，如果第二个字符是“u”说明是“push”，然后从第五个位置开始截取就可获得数字，同时将该数字入栈；否则是“pop”，将栈中的数字出栈。
  • 在得到先序和中序序列之后，从前往后遍历先序序列，每次从中序序列中找到左右子树的范围，使用left和right标记，进行递归，如果left！=right，说明当前结点不是叶子结点，继续递归左右子树；如果left==right，说明该结点是叶子结点。
  • 注意，其中的两个变量cur和printtag必须设成全局变量，因为使用该变量的函数中有递归函数，如果cur设成局部变量会发生段错误；如果printtag设成局部变量，会发生输出格式错误。

解题思路和代码参考https://blog.csdn.net/xyt8023y/article/details/47443489

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <vector>
 4 #include <stack>
 5 #include <string>
 6 #include <sstream>
 7 using namespace std;
 8 
 9 //分别存储先序序列和中序序列
10 vector<int>preorder;
11 vector<int>inorder;
12 
13 typedef struct TreeNode* Node;
14 struct TreeNode
15 {
16     int num;
17     Node left;
18     Node right;
19 };
20 
21 //根据结点值在中序序列中寻找根结点的下标位置
22 int FindRootIndex(int rootnum) {
23     for (unsigned int i = 0; i < inorder.size(); i++) {
24         if (inorder[i] == rootnum)
25             return i;
26     }
27     return -1;
28 }
29 
30 //通过给出中序序列中左右子树的范围建树
31 int cur;//定义遍历preorder的变量
32 Node CreateTree(int left, int right) {
33     //int cur = 0;
34     if (left > right)
35         return NULL;
36     int rootnum = preorder[cur];
37     cur++;
38     int rootIndex = FindRootIndex(rootnum);
39     //建立该根结点
40     Node T = (Node)malloc(sizeof(TreeNode));
41     T->num = rootnum;
42     T->left = NULL;
43     T->right = NULL;
44     if (left != right) {
45         T->left = CreateTree(left, rootIndex - 1);
46         T->right = CreateTree(rootIndex + 1, right);
47     }
48     return T;
49 }
50 
51 
52 //后序遍历
53 int printtag = 0;
54 void PostTraver(Node T) {
55     //int printtag = 0;
56     if (T) {
57         PostTraver(T->left);
58         PostTraver(T->right);
59         if (!printtag) {
60             cout << T->num;
61             printtag = 1;
62         }
63         else
64             cout << " " << T->num;
65     }
66 }
67 
68 int main() {
69     int n;
70     stack<int>s;
71     cin >> n;
72     getchar();
73     for (int i = 0; i < n*2; i++) {
74         string str;
75         getline(cin, str);
76 
77         if (str[1] == 'u') {
78             int num = stoi(str.substr(5));
79             preorder.push_back(num);
80             s.push(num);
81         }
82         else {
83             int num = s.top();
84             s.pop();
85             inorder.push_back(num);
86         }
87     }
88     Node T = CreateTree(0,inorder.size()-1);
89     PostTraver(T);
90     return 0;
91 }

 

posted @ 2020-04-22 17:22  PennyXia  阅读(192)  评论(0)  编辑  收藏  举报