03-树3 Tree 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 NN (\le 30≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to NN). Then 2N2N 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

题目大意：从push进去顺序可以看出是先序遍历，从pop进去的顺序可以看出是中序遍历，现在要求后序遍历。

思路：只要知道中序和其余任一个，就可以知道另外一个遍历。现在是中序和先序，即一棵树先序的是第一个是后序的最后一个，然后再通过中序找到其左右子树，重复上述操作。

#include<iostream>#include<string>#include<vector>#include<stack>using namespace std;int num;void getpostorder(vector<int> preorder,int preL,vector<int> inorder,int inL,    vector<int>& postorder,int postL,int n){    if (n == 0) return;    if (n == 1) {        postorder[postL] = preorder[preL];        return;    }     preorder[preL];    postorder[postL + n - 1] = preorder[preL];    //在中序遍历数组上找出root的位置    int i = 0;    while (i < n) {        if (inorder[inL + i] == preorder[preL]) break;        ++i;    }    int L = i, R = n - i - 1;    getpostorder(preorder, preL + 1, inorder, inL, postorder, postL, L);    getpostorder(preorder, preL + 1 + L, inorder, inL + L + 1, postorder, postL + L, R);}vector<vector<int>> getorder(){    cin >> num;    vector<int> preOrder(num, 0);    vector<int> inOrder(num, 0);    stack<int> st;    int pre = 0, in = 0;    string temp;    int data;    for (int i = 0; i < 2 * num; i++)    {        cin >> temp;        if (temp == "Push")        {            cin >> data;            st.push(data);            preOrder[pre++] = data;        }        else if (temp == "Pop")        {            inOrder[in++] = st.top();            st.pop();        }    }    return{ preOrder, inOrder };}int main(){    auto oredr = getorder();    vector<int> postorder(num, 0);    getpostorder(oredr[0], 0, oredr[1], 0, postorder, 0, num);    int i = 0;    for (; i < num-1; i++)    {        cout << postorder[i] << ' ';    }    cout << postorder[i];}