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

题意解析：入栈顺序即为先序遍历的顺序，出栈顺序即为中序遍历的顺序

解题思路：

    1.根据入栈出栈顺序，建立先序遍历数组与中序遍历数组。

    2.取先序序列中的第一个元素，该元素为根结点

    3.根据根结点在中序序列中查找根结点的位置，从而得到该树左子树结点个数（L）与右子树的结点个数（R）

    4.在后序序列数组中，第0到第L个元素为左子树，第L+1到第L+R个元素为右子树，最后一个元素为根结点

#include 
     
      #include 
      
       #include 
       
        using namespace std;#define MAXSIZE 30int preOrder[MAXSIZE];int inOrder[MAXSIZE];int postOrder[MAXSIZE];void Solve(int preL, int inL, int postL, int n);int main(){    for (int i = 0; i < MAXSIZE; i++)    //初始化数组    {        preOrder[i] = 0;        inOrder[i] = 0;        postOrder[i] = 0;    }    stack
        
          inputStack;    int nodeNum;    cin >> nodeNum;    int i, data;    int preIndex = 0, inIndex = 0, postIndex = 0;    string str;    for (i = 0; i < 2 * nodeNum; i++)    {        cin >> str;        if (str == "Push")        {            cin >> data;            preOrder[preIndex++] = data;            inputStack.push(data);        }        else if (str == "Pop")        {            inOrder[inIndex++] = inputStack.top();            inputStack.pop();        }    }    Solve(0, 0, 0, nodeNum);    for (int i = 0; i < nodeNum; i++)    {        if ( i == nodeNum - 1 )        {            cout << postOrder[i] << endl;        }        else        {            cout << postOrder[i] << ' ';        }    }}void Solve(int preL, int inL, int postL, int n){    if ( n == 0 )    {        return;    }    if ( n == 1 )    {        postOrder[postL] = preOrder[preL];        return;    }    int root = preOrder[preL];    postOrder[postL + n - 1] = root;    int L, R;    int i;    for (i = 0; i < n; i++)    {        if ( inOrder[inL + i] == root )        {            break;        }    }    L = i;    //左子树结点个数    R = n - L - 1;    //右子树结点个数    Solve(preL + 1, inL, postL, L);    Solve(preL + L + 1, inL + L + 1, postL + L, R);}