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.
题意:给出一棵BST树的先序序列,再给出m组顶点u,v,让你找出u和v的最近公共祖先,并按要求输出。
思路: 如果先建立BST树在通过DFS求最近公共祖先,则会有几组case超时,考虑到这是一棵BST树,且给定的是先序序列,有如下特点:对于任意给定的顶点u和v,从头开始遍历该先序序列,如果当前顶点a>=u并且a<=v或者a>=v且a<=u,则可以断定a是u和v的最近公共祖先,为了防止u或v不在序列中的情况,可以使用map在输入时,做一个映射,如果u或v不在map中则不是树中的顶点。
参考代码:
#include<cstdio>
#include<unordered_map>
using namespace std;
unordered_map<int,int> mp;
int n,m,u,v,a[10010];
int main()
{
scanf("%d%d",&m,&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
mp[a[i]]=1;
}
for(int k=0;k<m;k++){
scanf("%d%d",&u,&v);
if(mp.find(u)==mp.end()&&mp.find(v)==mp.end())
printf("ERROR: %d and %d are not found.\n",u,v);
else if(mp.find(u)==mp.end())
printf("ERROR: %d is not found.\n",u);
else if(mp.find(v)==mp.end())
printf("ERROR: %d is not found.\n",v);
else{
bool flag=false;
int LCA=0;
for(int i=0;i<n&&!flag;i++){
if((a[i]>=u&&a[i]<=v)||(a[i]>=v&&a[i]<=u)){
LCA=a[i];
flag=true;
}
}
if(LCA==u) printf("%d is an ancestor of %d.\n",u,v);
else if(LCA==v) printf("%d is an ancestor of %d.\n",v,u);
else printf("LCA of %d and %d is %d.\n",u,v,LCA);
}
}
return 0;
}