浙大数据结构pta——03-树3：Tree Traversals Again (25分)

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

这道题就是一道已知先序和中序然后让你用后序遍历树并输出，push的顺序即为先序，pop的顺序即为中序。那么就需要递归的寻找根节点，先序的第一个元素为根节点，那么中序中找到根节点，根节点的前半部分为左子树，后半部分为右子树，就这样不断递归，然后存入后序中即可，详见代码及注释

比如题目中的例子
先序：1 2 3 4 5 6
中序：3 2 4 1 6 5
后序：3 4 2 6 5 1

#include 
#include 
#include 
#include 

using namespace std;

void GetPostOrder(vector<int> &PreOrder/*先序数组*/,int PreL/*先序指针*/,vector<int> &InOrder/*中序数组*/,int InL/*中序指针*/,
                  vector<int> &PostOrder/*后序数组*/,int PostL/*后序指针*/,int n)
{
    if(n == 0)
    {
        return;
    }
    else if( n == 1)
    {
        PostOrder[PostL] = PreOrder[PreL];
        return;
    }
    int root = PreOrder[PreL];/*先序序列的第一个为根节点*/
    PostOrder[PostL + n - 1] = root;/*后序序列中最后访问根节点*/
    /*在中序序列中找到根节点*/
    int i = 0;
    while(i < n)
    {
        if(InOrder[InL + i] == root)
        {
            break;/*找到对应根节点*/
        }
        i++;
    }
    int L = i;/*左子树*/
    int R = n - i - 1;/*右子树*/
    GetPostOrder(PreOrder,PreL + 1,InOrder,InL,PostOrder,PostL,L);/*遍历左子树*/
    GetPostOrder(PreOrder,PreL + L + 1,InOrder,InL + L + 1,PostOrder,PostL + L,R);/*遍历右子树*/
}

int main()
{
    int n;
    cin >> n;
    vector<int> PreOrder(n,0);
    vector<int> InOrder(n,0);
    stack<int> st;
    int PreL = 0,InL = 0;
    for(int i = 0 ; i < 2 * n; i++)
    {
        string str;
        int temp;
        cin >> str;
        if(str == "Push")
        {
            cin >> temp;
            PreOrder[PreL++] = temp;/*先序序列*/
            st.push(temp);
        }
        else
        {
            InOrder[InL++] = st.top();/*中序序列*/
            st.pop();
        }
    }
    vector<int> PostOrder(n,0);
    GetPostOrder(PreOrder,0,InOrder,0,PostOrder,0,n);
    for(int i = 0 ; i < n ; i++)
    {
        if(i < n - 1)
        {
            cout << PostOrder[i] << " ";
        }
        else
        {
            cout << PostOrder[i] << endl;
        }
    }
    return 0;
}