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.
这一题是较为简单的lca问题,尽管最后还是有一个测试点超时了(仍然未解决)。这个问题的解决方法便是先手工实现以找到规律,而后采用代码的方法解出。他给出了这个序列是先序序列,而后这个先序序列节点的顺序是左孩子右孩子必须在根节点之后,而且左孩子比根大,右孩子比根小,(这读了好久的题才读懂)。而后,需要做的是找规律,这里找到的规律是其lca首先要在两节点的位置之前,其次,值要在两个节点大小之间。若在两点之前找不到对应的点,则两点是父子关系,出现在前的便是父亲节点。
因此,这里定了两个map(是不是麻烦了些?)一个是下标对数值,另一个是数值对下标,以解决最前公共子节点问题。第二个注意的地方是设置一个set,用来存放输入的数值。以此来确定节点是不是在树内。
完整代码如下(还有超时的一部分不知道怎么改):
#include <bits/stdc++.h>
using namespace std;
vector <int> pre;
set<int> s;
map<int,int>pos,pos2;
void pre_order(int c,int d)
{
int pos_c=pos2[c];
int pos_d=pos2[d];
int max_p=pos_c<pos_d?pos_c:pos_d;//max_p 用于找出c与d两点哪一个位置在前
int root=0;
int flag=0;
for(int i=1;i<max_p;i++)
{
if((pos[i]>c&&pos[i]<d)||(pos[i]<c&&pos[i]>d))
{
root=pos[i];
cout<<"LCA of "<<c<<" and "<<d<<" is "<<root<<"."<<endl;
flag=1;
break;
}
}
if(flag==0)
{
if(max_p==pos_c)
{
cout<<c<<" is an ancestor of "<<d<<"."<<endl;
}
else
{
cout<<d<<" is an ancestor of "<<c<<"."<<endl;
}
}
}
int main()
{
int n,m;
cin>>n>>m;
int i=0;
pre.resize(m+1);
for(i=1;i<=m;i++)
{
cin>>pre[i];
pos[i]=pre[i];
pos2[pre[i]]=i;
s.insert(pre[i]);
}
int c,d;
for(i=1;i<=n;i++)
{
cin>>c>>d;
if(s.find(c)==s.end()&&s.find(d)==s.end())
{
cout<<"ERROR: "<<c<<" and "<<d<<" are not found."<<endl;
}
else if(s.find(c)!=s.end()&&s.find(d)==s.end())
{
cout<<"ERROR: "<<d<<" is not found."<<endl;
}
else if(s.find(c)==s.end()&&s.find(d)!=s.end())
{
cout<<"ERROR: "<<c<<" is not found."<<endl;
}
else
{
pre_order(c,d);
}
}
return 0;
}