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
- set_new_handler
转自:http://www.cnblogs.com/hbt19860104/archive/2012/10/10/2717873.html 以及 http://zhaoweizhuanshuo.blo ...
- [2016北京集训试题6]魔法游戏-[博弈论-sg函数]
Description Solution 首先,每个节点上的权值可以等价于该节点上有(它的权的二进制位数+1)个石子,每次可以拿若干个石子但不能不拿. 然后就发现这和NIM游戏很像,就计算sg函数em ...
- c++ 创建二叉树
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <ctype.h> ...
- 5286: [Hnoi2018]转盘
5286: [Hnoi2018]转盘 链接 分析: $\min\limits_{i=1}^n \{ \max\limits_{j=i}^{i + n - 1} \{ a_{j}+i \} \} +n- ...
- 一个本地DNS解析和mysql授权导致的Mysq连接失败问题(Access denied for user 'loan'@'kfcsdb1' (using password: YES))
web:/home/web/ -u loan -p loan Enter password: ERROR 1045 (28000): Access denied for user 'loan'@'kf ...
- Scikit-Learn机器学习入门
现在最常用的数据分析的编程语言为R和Python.每种语言都有自己的特点,Python因为Scikit-Learn库赢得了优势.Scikit-Learn有完整的文档,并实现很多机器学习算法,而每种算法 ...
- 解决 java循环中使用 Map时 在put值时value值被覆盖的问题
其实很简单,只需要把容器换成list 然后在循环中,每次循环末尾map = new HashMap() 或者直接在循环中一开始就实例化hashmap(Map map = new HashMap();) ...
- python爬虫之解析库正则表达式
上次说到了requests库的获取,然而这只是开始,你获取了网页的源代码,但是这并不是我们的目的,我们的目的是解析链接里面的信息,比如各种属性 @href @class span 抑或是p节点里 ...
- Linux入门基础(六):Linux系统启动
系统启动流程 BIOS MBR : Boot Code 执行引导程序-GRUB 加载内核 执行init runlevel BIOS BIOS(basic input output system),我们 ...
- react.js插件开发,x-dailog弹窗浮层组件
react.js插件开发,x-dailog弹窗浮层组件 我认为,每一个组件都应该有他自带的样式和属性事件回调配置.所以我会给x-dialog默认一套简单的样式,和各种默认的配置项.所有react插件示 ...