PAT Advanced 1143 Lowest Common Ancestor (30) [二叉查找树 LCA]
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
- u,v在root两边,则root为u,v最近公共祖先节点
- u,v都在root左边,则最近公共祖先在root左子树,递归查找
- u,v都在root右边,则最近公共祖先在root右子树,递归查找
- u==root,则u是v的最近公共祖先节点
- 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
- PAT 甲级 1143 Lowest Common Ancestor
https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312 The lowest common ance ...
- 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 ...
- 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 ...
- [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 ...
- 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 ...
- 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 ...
- 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 ...
- [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 ...
- 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
- 吴裕雄 Bootstrap 前端框架开发——Bootstrap 按钮:分割按钮
<!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...
- InnoDB 和 MyISAM的索引区别
MyISAM索引实现 MyISAM索引文件和数据文件是分离的,索引文件的data域保存记录所在页的地址(物理存储位置),通过这些地址来读取页,进而读取被索引的行数据. MyISAM的索引原理图如下,C ...
- vscode修改样式
以修改上方滚动条宽度为例 打开开发者工具 help->toggle developer tool 或者快捷键 ctrl+shift+i 选择滚动条,找到css对应文件 鼠标移上去可以看到路径,类 ...
- 2-10 就业课(2.0)-oozie:12、cm环境搭建的基础环境准备
8.clouderaManager5.14.0环境安装搭建 Cloudera Manager是cloudera公司提供的一种大数据的解决方案,可以通过ClouderaManager管理界面来对我们的集 ...
- CentOS LVM 卷在线扩容
场景: vmware 虚拟机,装了CentOS ,更改了虚拟机磁盘的大小:从200G,扩展到320G,可以参考本文写了步骤. 1. 在线扫描虚拟机SCSI新增的容量 # for i in `find ...
- mongodb - 关联字段
1,博客表结构 Blog.js var mongoose = require('mongoose') mongoose.connect('mongodb://localhost/test',{ us ...
- Program-Language
1. 主流编程语言 2. 编程语言分类 2.1 编译or解释 2.2 按照客观系统的描述可分为两类 2.3 按照编程范型可分为 3. 语言范式 Paradigm 4. 计算机语 ...
- Windows系统批处理命令实现计划关机
操作步骤: 1.新建一个文本文件,粘贴下面代码,保存为shutdown.bat @echo off echo 请输入延迟关机分钟数 echo 小于1分钟将视为立即关机,负数为取消关机 set /p t ...
- vue2-dragula vue拖拽组件
https://github.com/kristianmandrup/vue2-dragula git 地址 https://github.com/kristianmandrup/vue2-dragu ...
- java 立方变自身
立方变自身 观察下面的现象,某个数字的立方,按位累加仍然等于自身. 1^3 = 1 8^3 = 512 5+1+2=8 17^3 = 4913 4+9+1+3=17 - 请你计算包括1,8,17在内, ...