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 = 100005;

int main(int argc, char *argv[]) {
    int m,n;
    int list[maxn];
    map<int,bool> mp;
    
    scanf("%d %d",&m,&n);
    for(int i=0;i<n;i++){
        scanf("%d",&list[i]);
        mp[list[i]]=true;
    }
    int a,b,p;
    for(int i=0;i<m;i++){
        scanf("%d %d",&a,&b);
        int flag=0;
        if(mp[a]==false&&mp[b]==false){
            flag=5;
        }else{
            if(mp[a]==false){
                flag=3;
            }else if(mp[b]==false){
                flag=4;
            }
        }
        if(flag==0){
            for(int i=0;i<n;i++){
                p=list[i];
                if((a<=p&&p<=b) || (b<=p&&p<=a)){
                    break;
                }
            }
        }
        if(flag==5){
            printf("ERROR: %d and %d are not found.
",a,b);
        }else if(flag==4){
            printf("ERROR: %d is not found.
",b);
        }else if(flag==3){
            printf("ERROR: %d is not found.
",a);
        }else if(p==a){
            printf("%d is an ancestor of %d.
",a,b);
        }else if(p==b){
            printf("%d is an ancestor of %d.
",b,a);
        }else{
            printf("LCA of %d and %d is %d.
",a,b,p);
        }
    }
}