题目

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

  • The lef subtree of a node contains only nodes with keys less than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
  • Both the lef and right subtrees must also be binary search trees.

Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line “LCA of U and V is A.” if the LCA is found and A is the key. But if A is one of U and V, print “X is an ancestor of Y.” where X is A and Y is the other node. If U or V is not found in the BST, print in a line “ERROR: U is not found.” or “ERROR: V is not found.” or “ERROR: U and V are not found.”.

Sample Input:

6 8

6 3 1 2 5 4 8 7

2 5

8 7

1 9

12 -3

0 8

99 99

Sample Output:

LCA of 2 and 5 is 3.

8 is an ancestor of 7.

ERROR: 9 is not found.

ERROR: 12 and -3 are not found.

ERROR: 0 is not found.

ERROR: 99 and 99 are not found.

题目分析

已知二叉查找树的前序序列,求每个测试样例中两个节点最近公共祖先节点

解题思路

利用二叉查找树特点:根节点,大于左子树所有结点,小于右子树所有结点

前序序列依次遍历,即为从左到右从下到上每个根节点,判断测试样例两个节点与每个根节点大小关系

  1. u,v在root两边,则root为u,v最近公共祖先节点
  2. u,v都在root左边,则最近公共祖先在root左子树,递归查找
  3. u,v都在root右边,则最近公共祖先在root右子树,递归查找
  4. u==root,则u是v的最近公共祖先节点
  5. v==root,则v是u的最近公共祖先节点

易错点

利用前序升序排序得到中序,前序+中序查找lca会有一个测试点超时(对比A1151)

Code

#include <iostream>
#include <vector>
#include <map>
using namespace std;
map<int,bool> mp;
int main(int argc,char * argv[]) {
int m,n,u,v,a;
scanf("%d %d",&m,&n);
vector<int> pre(n);
for(int i=0; i<n; i++) {
scanf("%d",&pre[i]);
mp[pre[i]]=true;
}
for(int i=0; i<m; i++) {
scanf("%d %d",&u,&v);
for(int i=0; i<n; i++) {
a=pre[i];
if(a>=u&&a<=v||a<=u&&a>=v)break;
}
if(mp[u]==false&&mp[v]==false) //都没找到
printf("ERROR: %d and %d are not found.\n",u,v);
else if(mp[u]==false||mp[v]==false)
printf("ERROR: %d is not found.\n",mp[u]==false?u:v);
else if(a==v||a==u)
printf("%d is an ancestor of %d.\n",a,a==v?u:v);
else
printf("LCA of %d and %d is %d.\n",u,v,a);
}
return 0;
}

PAT Advanced 1143 Lowest Common Ancestor (30) [二叉查找树 LCA]的更多相关文章

  1. PAT 甲级 1143 Lowest Common Ancestor

    https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 The lowest common ance ...

  2. 1143. Lowest Common Ancestor (30)

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  3. PAT甲级1143 Lowest Common Ancestor【BST】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 题意: 给定一个二叉搜索树,以及他的前 ...

  4. [PAT] 1143 Lowest Common Ancestor(30 分)

    1143 Lowest Common Ancestor(30 分)The lowest common ancestor (LCA) of two nodes U and V in a tree is ...

  5. PAT 1143 Lowest Common Ancestor[难][BST性质]

    1143 Lowest Common Ancestor(30 分) The lowest common ancestor (LCA) of two nodes U and V in a tree is ...

  6. PAT 1143 Lowest Common Ancestor

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  7. 1143 Lowest Common Ancestor

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...

  8. [PAT] 1143 Lowest Common Ancestor(30 分)1145 Hashing - Average Search Time(25 分)

    1145 Hashing - Average Search Time(25 分)The task of this problem is simple: insert a sequence of dis ...

  9. PAT_A1143#Lowest Common Ancestor

    Source: PAT A1143 Lowest Common Ancestor (30 分) Description: The lowest common ancestor (LCA) of two ...

随机推荐

  1. Golang编程的工程管理

    Golang编程的工程管理 作者:尹正杰 版权声明:原创作品,谢绝转载!否则将追究法律责任.

  2. 02.swoole学习笔记--UDP服务器

    <?php //创建服务器 $serv=,SWOOLE_PROCESS,SWOOLE_SOCK_UDP); //bool $swoole_server->on(string $event, ...

  3. Python连载61-tkinter三种布局

    一.pack布局举例 #pack布局案例 import tkinter baseFrame = tkinter.Tk() #以下代码都是创建一个组件,然后布局 btn1 = tkinter.Butto ...

  4. PHP mb_substr mbstring 函数

    定义和用法 mb_substr - 获取部分字符串 版本支持 PHP4 PHP5 PHP7 支持 支持 支持 5.4.8 length 传入 NULL,则从 start 提取到字符串的结尾处. 在之前 ...

  5. JS:递归基础及范例——斐波那契数列 、 杨辉三角

    定义:程序调用自身的编程技巧称为递归.一个过程或函数在其定义或说明中有直接或间接调用自身的一种方法,它通常把一个大型复杂的问题层层转化为一个与原问题相似的规模较小的问题来求解,递归策略只需少量的程序就 ...

  6. manjaro更换深度桌面 卸载ked桌面

    下载需要的桌面环境 pacman -S deepin deepin-extra 修改/etc/lightdm/lightdm.conf sudo nano /etc/lightdm/lightdm.c ...

  7. 吴裕雄 Bootstrap 前端框架开发——Bootstrap 字体图标(Glyphicons):glyphicon glyphicon-signal

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta name ...

  8. 自制spring中bean加载机制,仅做笔记自用

  9. nodejs 杂七杂八

    nodejs => 提供核心模块语法 node中的回调函数 都是异步

  10. 面向对象设计思想和MVC设计模式

    虽然之前学习Java时有接触过面向对象的设计思想,但因当时Java没学好.所以导致这两天讲php的面向对象设计时,感到没有头绪,这也反应了我练习少和逻辑能力的不足.而MVC设计思想 面向对象就是要将系 ...