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