PAT 甲级 1151 LCA in a Binary Tree
https://pintia.cn/problem-sets/994805342720868352/problems/1038430130011897856
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.
Given any two nodes in a binary tree, 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 binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. 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 binary tree, 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
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99
Sample Output:
LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.
代码:
#include <bits/stdc++.h>
using namespace std; int N, M;
vector<int> in, pre;
map<int, int> pos; void lca(int inl, int inr, int preroot, int a, int b) {
if(inl > inr) return ;
int inroot = pos[pre[preroot]], ain = pos[a], bin = pos[b];
if(ain < inroot && bin < inroot)
lca(inl, inroot - 1, preroot + 1, a, b);
else if((ain > inroot && bin < inroot) || (ain < inroot && bin > inroot))
printf("LCA of %d and %d is %d.\n", a, b, in[inroot]);
else if(ain > inroot && bin > inroot)
lca(inroot + 1, inr, preroot + 1 + inroot - inl, a, b);
else if(ain == inroot)
printf("%d is an ancestor of %d.\n", a, b);
else if(bin == inroot)
printf("%d is an ancestor of %d.\n", b, a);
} int main() {
scanf("%d%d", &M, &N);
in.resize(N + 1), pre.resize(N + 1);
for(int i = 1; i <= N; i ++) {
scanf("%d", &in[i]);
pos[in[i]] = i;
}
for(int i = 1; i <= N; i ++)
scanf("%d", &pre[i]); for(int i = 0; i < M; i ++) {
int x, y;
scanf("%d%d", &x, &y);
if(pos[x] == 0 && pos[y] == 0)
printf("ERROR: %d and %d are not found.\n", x, y);
else if(pos[x] == 0 || pos[y] == 0)
printf("ERROR: %d is not found.\n", pos[x] == 0 ? x : y);
else
lca(1, N, 1, x, y);
}
return 0;
}
LCA
FHFHFH
PAT 甲级 1151 LCA in a Binary Tree的更多相关文章
- PAT甲级|1151 LCA in a Binary Tree 先序中序遍历建树 lca
给定先序中序遍历的序列,可以确定一颗唯一的树 先序遍历第一个遍历到的是根,中序遍历确定左右子树 查结点a和结点b的最近公共祖先,简单lca思路: 1.如果a和b分别在当前根的左右子树,当前的根就是最近 ...
- PAT Advanced 1151 LCA in a Binary Tree (30) [树的遍历,LCA算法]
题目 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both ...
- PAT 1151 LCA in a Binary Tree[难][二叉树]
1151 LCA in a Binary Tree (30 分) The lowest common ancestor (LCA) of two nodes U and V in a tree is ...
- 【PAT 甲级】1151 LCA in a Binary Tree (30 分)
题目描述 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has bo ...
- 1151 LCA in a Binary Tree(30 分)
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...
- 1151 LCA in a Binary Tree
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U ...
- 1151 LCA in a Binary Tree (30point(s))
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】1102 Invert a Binary Tree (25 分)(层次遍历和中序遍历)
题意: 输入一个正整数N(<=10),接着输入0~N-1每个结点的左右儿子结点,输出这颗二叉树的反转的层次遍历和中序遍历. AAAAAccepted code: #define HAVE_STR ...
- PAT_A1151#LCA in a Binary Tree
Source: PAT A1151 LCA in a Binary Tree (30 分) Description: The lowest common ancestor (LCA) of two n ...
随机推荐
- BZOJ4145_The Prices_KEY
题目传送门 看到M<=16经典状态压缩的数据范围,考虑题目. 一道类似于背包的题目. 设f[i][j]表示前i个商店,物品购买状态为j. 先将f[i][j]加上w[i](到i的路费),转移一次, ...
- PostreSQL崩溃试验全记录
磨砺技术珠矶,践行数据之道,追求卓越价值 回到上一级页面: PostgreSQL基础知识与基本操作索引页 回到顶级页面:PostgreSQL索引页 [作者 高健@博客园 luckyjackg ...
- 苏州Uber优步司机奖励政策(4月24日)
滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...
- 关于 Git SSH 使用的项目实践
Git 是分布式的代码管理工具,远程的代码管理是基于 SSH 的,所以要使用远程的 git 则需要SSH的配置. 一.简述 访问 Git 仓库可以使用 SSH Key 的方式,首先需要生成 Key. ...
- vue 组件间的通信
(1)props:用于父组件向子组件传递消息 使用方法: 在父组件中,使用子组件时,<Child v-bind:data="data"/>,通过v-bind把子组件需要 ...
- gradle springboot 项目运行的三种方式
一.java -jar 二.eclipse中 Java Application 三.命令行 gradle bootRun
- python图像处理(1)图像的打开与保存
使用python进行图像处理时有三种库可以使用分别是:PIL.matplotlib.pyplot.opencv(opencv未接触) 注意:matplotlib读取进来的图片是unit8,0-255范 ...
- 利用workbench对linux/Ubuntu系统中的mysql数据库进行操作
在上一篇文章中,我分享了在linux中如何安装mysql数据库,但是这只是安装了mysql的服务,并没有图形化管理界面,所以这样子操作起来并没有那么方便,那么现在我们就来实现如何利用在window中安 ...
- python快速入门——进入数据挖掘你该有的基础知识
这篇文章是用来总结python中重要的语法,通过这些了解你可以快速了解一段python代码的含义 Python 的基础语法来带你快速入门 Python 语言.如果你想对 Python 有全面的了解请关 ...
- eclipse检出SVN代码的详细流程
1.添加SVN资源库位置(未安装SVN,请先安装SVN) 2.因为该项目不是maven项目 所以还需要加入jar包(将项目lib里面的jar都Buile Path) 3.我这个项目需要修改编码格式 右 ...