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 (≤) 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

#include <cstdio>
#include <stack>
#include <cstring>
#include <cstdlib>
#include <iostream> 
using namespace std;
const int maxn = 32;
const int smaxn = 10;



int pre[maxn], in[maxn], post[maxn];
void input(int ) ;
int str2num(char s[], int start);
void solve(int postL, int preL, int inL, int n ) ;


int main() {
    int n;
    scanf("%d", &n);
    input(n);
    solve(0, 0, 0, n);
    int flag = 1;
    for(int i=0; i<n; i++) {
        if(flag == 1) {
            printf("%d", post[i]);
            flag = 0;
        }
        else printf(" %d", post[i]);
    }
    system("pause");
    return 0;
} 

void input(int n) {
    stack<int> st;
    char str[smaxn];
    int preIndex = 0, inIndex = 0;
    getchar();
    for(int i=0; i<2*n; i++) {
        cin.getline(str, maxn);
        if(strlen(str) != 3) {
            int tmp = str2num(str, 5);
            pre[preIndex++] = tmp ;
            st.push(tmp);
        }
        else {
            in[inIndex++] = st.top();
            st.pop();
        }
    }

}

int str2num(char s[], int start) 
{
    int sum = 0;
    for(; s[start]; start++) {
        sum = sum*10 + s[start] - '0';
    }
    return sum;
}

void solve(int postL, int preL, int inL, int n ) {

    if(n == 0) {
        return;
    }
    int root = pre[preL];
    int l_num, r_num, i;
    post[postL+n-1] = root;
    
    for(i=0; i<n; i++) {
        if(in[inL+i] == root) break;
    }
    l_num = i; r_num = n - i - 1;
    
    solve(postL, preL+1, inL, l_num);
    solve(postL+l_num, preL+l_num+1, inL+i+1, r_num);
    

}