1143 Lowest Common Ancestor

Posted 2021-01-17 lokwongho

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）

 1 #include <iostream>
 2 #include <map>
 3 using namespace std;
 4 const int maxn = 100005;
 5 
 6 int main(int argc, char *argv[]) {
 7     int m,n;
 8     int list[maxn];
 9     map<int,bool> mp;
10     
11     scanf("%d %d",&m,&n);
12     for(int i=0;i<n;i++){
13         scanf("%d",&list[i]);
14         mp[list[i]]=true;
15     }
16     int a,b,p;
17     for(int i=0;i<m;i++){
18         scanf("%d %d",&a,&b);
19         int flag=0;
20         if(mp[a]==false&&mp[b]==false){
21             flag=5;
22         }else{
23             if(mp[a]==false){
24                 flag=3;
25             }else if(mp[b]==false){
26                 flag=4;
27             }
28         }
29         if(flag==0){
30             for(int i=0;i<n;i++){
31                 p=list[i];
32                 if((a<=p&&p<=b) || (b<=p&&p<=a)){
33                     break;
34                 }
35             }
36         }
37         if(flag==5){
38             printf("ERROR: %d and %d are not found.
",a,b);
39         }else if(flag==4){
40             printf("ERROR: %d is not found.
",b);
41         }else if(flag==3){
42             printf("ERROR: %d is not found.
",a);
43         }else if(p==a){
44             printf("%d is an ancestor of %d.
",a,b);
45         }else if(p==b){
46             printf("%d is an ancestor of %d.
",b,a);
47         }else{
48             printf("LCA of %d and %d is %d.
",a,b,p);
49         }
50     }
51 }