• PTA 5-5 Tree Traversals Again (25)


    题目:http://pta.patest.cn/pta/test/16/exam/4/question/667

    PTA - Data Structures and Algorithms (English) - 5-5

    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.

    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   //即后序遍历树的结果

    解法转自:http://www.cnblogs.com/clevercong/p/4177802.html

    题目分析:

    动态的“添加结点”并遍历过程,不需要建树。

    借助“栈”进行树的后续遍历;( times:可由此栈操作判断中当前结点有左子树还是右子树中 )

    每次PUSH,times = 1;// PUSH:可看作访问左孩子

    每次POP检查栈顶记录的times:

    - 如果是1,弹出来变成2压回栈;// POP:可看作由左孩子转到访问右孩子

    - 如果是2,则弹出,放入存放结果的vector中,重复这一过程,直到栈顶times为1。// POP:可看作访问根节点

    所有PUSH与POP操作执行完毕时,输出vector内的数据和stack中的数据。 注意要处理最后的空格。

    注:此方法可推广到n叉树,只需改变times的值即可。

    image     imageimage       image……

    代码:

    #include <iostream>
    #include <stack>
    #include <vector>
    using namespace std;
    typedef struct node
    {
        int data;
        int times;
        node(int d, int t)
            :data(d), times(t)
        {};
    } Node;
    
    int main()
    {
        int n;
        cin >> n;
        string cmd;  //!PUSH or POP
        int x;
        stack<Node> sta;  //!栈:用来实现后序遍历
        vector<int> vec;  //!存放结果
        for(int i=0; i<2*n; i++)
        {
            cin >> cmd;
            if(cmd == "Push")
            {
                cin >> x;
                sta.push(Node(x, 1));
            }
            if(cmd == "Pop")
            {
                Node node = sta.top();
                sta.pop();
                if(node.times == 1)  //!times==1:弹出来变成2压回栈
                {
                    node.times = 2;
                    sta.push(node);
                }
                else if(node.times == 2)  //!times==2:弹出,放入存放结果的vector中
                {
                    vec.push_back(node.data);
                    while(sta.top().times == 2)  //!重复这一过程,直到栈顶的times==1
                    {
                        vec.push_back(sta.top().data);
                        sta.pop();
                    }
                    if (sta.size() != 0)  //!操作结束后栈还不为空,使栈顶的times=2
                    {
                        sta.top().times=2;                    
                    }
                }
            }
        }
        for(unsigned int i=0; i<vec.size(); i++)
        {
            cout << vec[i]<< " ";
        }
        while(sta.size() != 0)  //!PUSH/POP结束后栈中还有元素,直接pop()即可
        {
            cout << sta.top().data;
            sta.pop();
            if(sta.size() != 0)
                cout << " ";
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/claremore/p/4806304.html
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