PATA1086题解

通过前序遍历和中序遍历求树的后序遍历

//
//  main.cpp
//  PATA1086
//
//  Created by Phoenix on 2018/2/19.
//  Copyright © 2018年 Phoenix. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <stack>
using namespace std;
int pre[40], in[40], post[40];
int n;
struct node {
    int data;
    node *lchild, *rchild;
};

node* newnode(int x) {
    node* root = new node;
    root->data = x;
    root->lchild = root->rchild = NULL;
    return root;
}

node* create(int preL, int preR, int inL, int inR) {
    if(preL > preR) return NULL;
    node* root = newnode(pre[preL]);
    //printf("%d ", pre[preL]);
    int k;
    for(int i = inL; i <= inR; i++) {
        if(in[i] == pre[preL]) k = i;
    }
    root->lchild = create(preL + 1, preL + k - inL, inL, k - 1);
    root->rchild = create(preL + k - inL + 1, preR, k + 1, inR);
    return root;
}

int num = 0;
void postorder(node* root) {
    if(root == NULL) return;
    postorder(root->lchild);
    postorder(root->rchild);
    printf("%d", root->data);
    if(num++ < n - 1) printf(" ");
    else printf("\n");
}

int main(int argc, const char * argv[]) {
    int num1 = 0, num2 = 0;
    scanf("%d", &n);
    stack<int> st;
    for(int i = 0; i < 2 * n; i++) {
        char s[5];
        scanf("%s", s);
        if(s[1] == 'u') {
            scanf("%d", &pre[num1]);
            st.push(pre[num1++]);
        } else {
            int top = st.top();
            st.pop();
            in[num2++] = top;
        }
    }
    node* root = create(0, n - 1, 0, n - 1);
    postorder(root);
    return 0;
}