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 and V as descendants.
A binary search tree (BST) is recursively defined as a binary tree which has the following properties:
- The left 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 left 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 (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), 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.
.
知识点:搜索二叉树
思路:不用建BST树!将输入的数字按顺序存在list[]中;对于每个test,遍历一遍数组,将当前结点标记为a,如果u和v分别在a的左、右,或者u、v其中一个就是当前a,即(a >= u && a <= v) || (a >= v && a <= u),说明找到了这个共同最低祖先a,退出当前循环~最后根据要求输出结果即可。
因为输入的错误数字的范围不确定,所以要用map来判断输入是否合法
(看了 liuchuo.net 的思路)
#include <iostream>
#include <map>
using namespace std;
const int maxn = ; int main(int argc, char *argv[]) {
int m,n;
int list[maxn];
map<int,bool> mp; scanf("%d %d",&m,&n);
for(int i=;i<n;i++){
scanf("%d",&list[i]);
mp[list[i]]=true;
}
int a,b,p;
for(int i=;i<m;i++){
scanf("%d %d",&a,&b);
int flag=;
if(mp[a]==false&&mp[b]==false){
flag=;
}else{
if(mp[a]==false){
flag=;
}else if(mp[b]==false){
flag=;
}
}
if(flag==){
for(int i=;i<n;i++){
p=list[i];
if((a<=p&&p<=b) || (b<=p&&p<=a)){
break;
}
}
}
if(flag==){
printf("ERROR: %d and %d are not found.\n",a,b);
}else if(flag==){
printf("ERROR: %d is not found.\n",b);
}else if(flag==){
printf("ERROR: %d is not found.\n",a);
}else if(p==a){
printf("%d is an ancestor of %d.\n",a,b);
}else if(p==b){
printf("%d is an ancestor of %d.\n",b,a);
}else{
printf("LCA of %d and %d is %d.\n",a,b,p);
}
}
}
1143 Lowest Common Ancestor的更多相关文章
- 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(30 分)
1143 Lowest Common Ancestor(30 分)The lowest common ancestor (LCA) of two nodes U and V in a tree is ...
- 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 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 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 ...
- 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)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 ...
- [LeetCode] Lowest Common Ancestor of a Binary Tree 二叉树的最小共同父节点
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According ...
随机推荐
- Django 实现登陆验证码
一 基本使用方法 Python生成随机验证码,需要使用PIL模块 安装: pip3 install pillow 基本使用 1 创建图片 from PIL import Image, ImageDra ...
- Android.Libraries
1. Android Dependencies, Referenced Libraries, Android Private Libraries Android Private Libraries - ...
- 洛谷3119 [USACO15JAN]草鉴定Grass Cownoisseur
原题链接 显然一个强连通分量里所有草场都可以走到,所以先用\(tarjan\)找强连通并缩点. 对于缩点后的\(DAG\),先复制一张新图出来,然后对于原图中的每条边的终点向新图中该边对应的那条边的起 ...
- Git 分支 主干
~/Desktop/work/movies/movie(apps) $ git status //先查看是否有需要提交的东西# On branch appsnothing to commit (wo ...
- 小话C源码移植
我们知道国外很多程序员工作在linux / unix 环境下,所以有很多优秀的c/c++语言代码不能直接在windows平台进行编译. 很多时候我们只能使用msys, cmake等工具进行模拟环境编译 ...
- uuid唯一吗
是唯一的.我在几台硬件完全相同(同一批购买的).软件也完全相同(用同一个GHOST系统安装)的电脑上试过: 不同的电脑上,wmic csproduct get uuid 获取的UUID码是不同的.另 ...
- Virtual Machine Kernel Panic : Not Syncing : VFS : Unable To Mount Root FS On Unknown-Block (0,0)
Virtual Machine Kernel Panic : Not Syncing : VFS : Unable To Mount Root FS On Unknown-Block (0,0) 33 ...
- flask学习视频
https://study.163.com/course/courseMain.htm?courseId=1004091002 主要 https://www.cnblogs.com/senlinyan ...
- HttpURLConnection 返回汉字乱码(全是问号)
public static String doPost(String urlStr, Map<String, Object> paramMap) throws Exception { UR ...
- 【转】你知道C#中的Lambda表达式的演化过程吗?
[转]你知道C#中的Lambda表达式的演化过程吗? 那得从很久很久以前说起了,记得那个时候... 懵懂的记得从前有个叫委托的东西是那么的高深难懂. 委托的使用 例一: 什么是委托? 个人理解:用来传 ...