zoukankan      html  css  js  c++  java
  • PAT A1138 Postorder Traversal (25 分)——大树的遍历

    Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder traversal sequence of the corresponding binary tree.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 50,000), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree.

    Sample Input:

    7
    1 2 3 4 5 6 7
    2 3 1 5 4 7 6
    

    Sample Output:

    3
    
     
     1 #include <stdio.h>
     2 #include <iostream>
     3 const int maxn=50010;
     4 struct node{
     5   int data;
     6   node* left,*right;
     7 };
     8 int in[maxn],pre[maxn];
     9 node* create(int prel,int prer,int inl,int inr){
    10   if(prel>prer) return NULL;
    11   node* root = new node;
    12   root->data = pre[prel];
    13   int k;
    14   for(k=inl;k<=inr;k++){
    15     if(in[k]==pre[prel]) break;
    16   }
    17   int numleft = k-inl;
    18   root->left = create(prel+1,prel+numleft,inl,k-1);
    19   root->right = create(prel+numleft+1,prer,k+1,inr);
    20   return root;
    21 }
    22 int main(){
    23   int n;
    24   scanf("%d",&n);
    25   for(int i=0;i<n;i++){
    26     int d;
    27     scanf("%d",&d);
    28     pre[i]=d;
    29   }
    30   for(int i=0;i<n;i++){
    31     int d;
    32     scanf("%d",&d);
    33     in[i]=d;
    34   }
    35   int prel=0,prer=n-1,inl=0,inr=n-1;
    36   int numleft=n;
    37   int k;
    38   while(numleft!=1){
    39         for(k=inl;k<=inr;k++){
    40             if(in[k]==pre[prel]) break;
    41         }
    42         int nl=k-inl;
    43         int nr=inr-k;
    44         if(nl==0){
    45             prel=prel+nl+1;
    46             prer=prer;
    47             inl=k+1;
    48             inr=inr;
    49             numleft=nr;
    50         }
    51         else{
    52             prel=prel+1;
    53             prer=prel+nl;
    54             inl=inl;
    55             inr=k-1;
    56             numleft=nl;
    57         }
    58    }
    59    printf("%d",in[inl]);
    60 }
    View Code

    注意点:这道题虽然给的时间很多650ms,但直接建树再查找还是会超时,好像可以用引用来避免超时。

    不建树其实就是通过两个序列来找到第一个叶子节点,当一个节点的左边或右边只有一个元素时,剩下的那个元素就是要输出的值。

    ---------------- 坚持每天学习一点点
  • 相关阅读:
    ES5 ES6 作用域声明部分
    js 内建函数reduce
    $apply的使用与否
    得分-星星
    CSS3中translate、transform和translation的区别和联系
    vue 学习笔记
    -webkit-line-clamp 多行文字溢出...
    八位二进制数为什么表示范围(-128~~+127)理解
    vs2017_enterprise正式版离线安装包bt下载
    RSA密钥之C#格式与Java格式转换
  • 原文地址:https://www.cnblogs.com/tccbj/p/10429516.html
Copyright © 2011-2022 走看看