1143 Lowest Common Ancestor （30 分）
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：
给你一棵二叉排序树的先序序列，求每次询问的LCA（最近公共祖先）
思路：
题目已知是一棵二叉排序树，其实将每个结点排序后得出的即是中序遍历，即已知中序和前序，建树，然后求LCA
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=10005;
int m,n;
struct node
{
int data;
node *lchild,*rchild;
};
bool cmp(int a,int b)
{
return a<b;
}
int pre[maxn];
int in[maxn];
node *build(int prel,int prer,int inl,int inr)
{
if(inl>inr)
return NULL;
int p1,p2;
node *root=new node;
root->data=pre[prel];
p1=inl;
while(in[p1]!=pre[prel])
p1++;//p1是根的位置
p2=p1-inl;
root->lchild=build(prel+1,prel+p2,inl,p1-1);
root->rchild=build(prel+p2+1,prer,p1+1,inr);
return root;
}
node *lca(node *root,int u,int v)
{
if(root==NULL)
return NULL;
if(root->data==u||root->data==v)
return root;
node *left=lca(root->lchild,u,v);
node *right=lca(root->rchild,u,v);
if(left&&right)
return root;
return left==NULL?right:left;
}
bool find(int u)//找是否存在于序列中
{
for(int i=0;i<n;i++)
{
if(u==pre[i])
return true;
}
return false;
}
int main()
{
scanf("%d%d",&m,&n);
for(int i=0;i<n;i++)
{
scanf("%d",&pre[i]);
in[i]=pre[i];
}
sort(in,in+n,cmp);
//这样就得到了层次遍历和前序遍历
node *root=build(0,n-1,0,n-1);
while(m--)
{
int u,v;
scanf("%d%d",&u,&v);
if(find(u)==false&&find(v)==false)
printf("ERROR: %d and %d are not found.\n",u,v);
else if(find(u)==false||find(v)==false)
printf("ERROR: %d is not found.\n",find(u)==false?u:v);
else
{
node *ans=lca(root,u,v);
if(ans->data==u||ans->data==v)
printf("%d is an ancestor of %d.\n",ans->data==u?u:v,ans->data==u?v:u);
else
printf("LCA of %d and %d is %d.\n",u,v,ans->data);
}
}
}