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..
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.题目大意：给出一棵二叉搜索树的前序遍历序列，查询给出值 u 和 v 的共同最低祖先的值。注意输出要求。
分析：先用一个 map 记录出现过的值，这样在读入查询的时候就能检查查询值是否在二叉搜索树中。这道题无需建树，不妨设 u 小于 v，且一定都存在于二叉搜索树中。u 和 v 的最近公共祖先要么是 u 或者 v 本身，要么是一个处于 u 和 v 中间的值。
查找最近公共祖先时，从序列的第一个值val 开始检查（也就是二叉搜索树的根节点），如果 val >= u 且 val <= v，说明就是要找的公共祖先，否则如果 val > u 且 val > v，说明 u 和 v 都在当前节点的右子树，则在前序序列中找到第一个大于 val 的值，继续查找；如果 val < u 且 val < v，说明 u 和 v 都在当前节点的左子树，则在前序序列中找到第一个小于 val 的值，继续查找。这样直到找到 u 或者 v 本身，或者是找到一个 val 处于 u 和 v 中间即为答案。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0x3fffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
#define NUMBER_OF_THREADS 10
using namespace std;
int find_anc(int nums[],int n,int index,int a,int b)
{
if(nums[index]>=a&&nums[index]<=b)return nums[index];
if(nums[index]>a&&nums[index]>b)return find_anc(nums,n,index+1,a,b);
else if(nums[index]<a&&nums[index]<b)
{
int ind=index+1;
for(;ind<n;++ind)
{
if(nums[ind]>=a||nums[ind]>=b)return find_anc(nums,n,ind,a,b);
}
}
return -1;
}
int main(void)
{
#ifdef test
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
clock_t start=clock();
#endif //test
int m,n,times=1;scanf("%d%d",&m,&n);
int nums[n+5]={0};
map<int,int>mp;
for(int i=0;i<n;++i)
{
scanf("%d",&nums[i]);
if(!mp[nums[i]])mp[nums[i]]=times,times++;
}
for(int i=0;i<m;++i)
{
int a,b;scanf("%d%d",&a,&b);
int f1=1,f2=1;
if(!mp[a])f1=0;
if(!mp[b])f2=0;
if(f1==0&&f2)printf("ERROR: %d is not found.\n",a);
else if(f1&&f2==0)printf("ERROR: %d is not found.\n",b);
else if(f1==0&&f2==0)printf("ERROR: %d and %d are not found.\n",a,b);
else
{
int ans=0;
if(a>b)ans=find_anc(nums,n,0,b,a);
else ans=find_anc(nums,n,0,a,b);
if(ans!=a&&ans!=b)printf("LCA of %d and %d is %d.\n",a,b,ans);
else if(ans==a)
{
printf("%d is an ancestor of %d.\n",a,b);
}
else if(ans==b)
{
printf("%d is an ancestor of %d.\n",b,a);
}
}
}
#ifdef test
clockid_t end=clock();
double endtime=(double)(end-start)/CLOCKS_PER_SEC;
printf("\n\n\n\n\n");
cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位
cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位
#endif //test
return 0;
}