PAT A1086

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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.
请输入图片描述

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

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#include "stdio.h"
//#include "math.h"
#include "string.h"
//#include "iostream"
//#include "stdlib.h"
#include "vector"
//#include "set"
//#include "map"
#include "stack"
//#include "queue"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn = 50;
struct node{
    int data;
    node* lchild;
    node* rchild;
};
int pre[maxn], in[maxn], post[maxn];//先序，中序以及后序
int n;//结点个数

//当前二叉树的先序序列区间为[preL, preR],中序序列区间为[inL, inR]
//create函数返回根节点
node* create(int preL, int preR, int inL, int inR){
    if (preL > preR) return NULL;
    node* root = new node;
    root->data = pre[preL];
    int k;
    for (k = inL; k <= inR; k++) {
        if (in[k] == pre[preL]) break;
    }
    int numLeft = k - inL;
    root->lchild = create(preL + 1, preL + numLeft, inL, k - 1);
    root->rchild = create(preL + numLeft + 1, preR, k + 1, inR);
    return root;
}
int num = 0;//已输出的结点个数
void postorder(node* root){
    //后序遍历
    if (root == NULL) {
        return;
    }
    postorder(root->lchild);
    postorder(root->rchild);
    printf("%d", root->data);
    num++;
    if(num < n) printf(" ");
}
int main(){
    scanf("%d", &n);
    char str[5];
    stack<int> st;
    int x, preIndex = 0, inIndex = 0;//入栈元素、先序序列位置及中序序列位置
    for (int i = 0; i < 2 * n; i++) {
        scanf("%s", str);
        if (strcmp(str, "Push") == 0) {//入栈
            scanf("%d", &x);
            pre[preIndex++] = x;
            st.push(x);
        }else{
            in[inIndex++] = st.top();
            st.pop();
        }
    }
    node* root = create(0, n-1, 0, n-1);
    postorder(root);
    return 0;
}