PAT 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.

思路：先建树再从上到下搜索

程序：

#include <iostream>
#include <cstdio>
#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;
struct node
{
  int val;
  struct node* left;
  struct node* right;
};

node* root = NULL;
vector<int> vec;
void createBST(node* &root,vector<int> &pre,vector<int> &mid,int pl,int pr,int ml,int mr)
{
  if(pr < pl || mr < ml) return;
  int pos = ml;
  while(mid[pos] != pre[pl])
  {
    pos++;
  }
  root = new node;
  root->val = pre[pl];
  root->left = NULL;
  root->right = NULL;
  createBST(root->left,pre,mid,pl+1,pl+pos-ml,ml,pos-1);
  createBST(root->right,pre,mid,pl+pos-ml+1,pr,pos+1,mr);
}
/*
 一开始的建树方法,一个节点一个节点地插到树上,有两个测试点超时
 void createBST(node* &root,int val)
 {
    if(root == NULL)
    {
       root = new node;
      root->val = val;
      root->left = NULL;
      root->right = NULL;
  }
  else
  {
  if(val > root->val)
    createBST(root->right,val);
  else
    createBST(root->left,val);
}
}
*/
node* findLCA(node* root,int u,int v)
{
  if((u >= root->val && v <= root->val) || (u <= root->val && v >= root->val))
    return root;
  else
  {
    if(u < root->val && v < root->val)
      return findLCA(root->left,u,v);
    else  if(u > root->val && v > root->val)
      return findLCA(root->right,u,v);
  }
}
int main()
{
  int n,m;
  scanf("%d%d",&n,&m);
  unordered_map<int,bool> map;
  vector<int> mid;
  for(int i = 0; i < m; i++)
  {
    int val;
    scanf("%d",&val);
    vec.push_back(val);
    map[val] = true;
  }
  mid = vec;
  sort(mid.begin(),mid.end());
  createBST(root,vec,mid,0,m-1,0,m-1);
  for(int i = 0; i < n; i++)
  {
    int u,v;
    scanf("%d%d",&u,&v);
    if(!map[u] && !map[v])
      printf("ERROR: %d and %d are not found.\n",u,v);
    else if(!map[u])
      printf("ERROR: %d is not found.\n",u);
    else if(!map[v])
      printf("ERROR: %d is not found.\n",v);
    else
    {
      node* res = findLCA(root,u,v);
      if(res->val == u)
      {
        printf("%d is an ancestor of %d.\n",u,v);
      }
      else if(res->val == v)
      {
        printf("%d is an ancestor of %d.\n",v,u);
      }
      else
      {
        printf("LCA of %d and %d is %d.\n",u,v,res->val);
      }
    }
  }
  return 0;
}