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.因为查询的数据里有负数，所以要用map来定义一个哈希表
因为根据二叉查找树的性质
左子树上的数据均小于等于根结点， 右子树均大于等于根结点。
题目给出的是先序遍历，即先访问根结点->左子树->右子树。
查询时只要遍历先序序列根据题目要求做条件判断即可
#include<cstdio>
#include<fstream>
#include<vector>
#include<map>
using namespace std;
vector<int> pre;
map<int, bool> hashtable;
int main(){
int m, n, a;
// freopen("d://in.txt","r",stdin);
scanf("%d%d", &m, &n);
for(int i=0; i<n; i++){
scanf("%d", &a);
pre.push_back(a);
hashtable[pre[i]]=true;
}
int u, v, x;
for(int i=0; i<m; i++){
scanf("%d%d", &u, &v);
for(int j=0; j<n; j++){
x=pre[j];
if((x>=u && x<=v) || (x<=u && x>=v)) break;
}
if(hashtable[u]==0 && hashtable[v]==0){
printf("ERROR: %d and %d are not found.\n", u, v);
} else if(hashtable[u]==0){
printf("ERROR: %d is not found.\n", u);
} else if(hashtable[v]==0){
printf("ERROR: %d is not found.\n", v);
} else if(x==u){
printf("%d is an ancestor of %d.\n", u, v);
} else if(x==v){
printf("%d is an ancestor of %d.\n", v, u);
} else{
printf("LCA of %d and %d is %d.\n", u, v, x);
}
}
return 0;
}