1143 Lowest Common Ancestor (30)

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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), 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

题目大意：给出一棵树的前序遍历，找出给出查找对的最近祖先；
思路：1.先用set记录出现过的节点
　　　2.通过遍历前序遍历来找公共祖先
     引用前序遍历是先遍历左子树， 再遍历右子树； 节点的值在给出的两个节点值之间， 则表示该点为两个节点的最近公共祖先

#include<iostream>
#include<vector>
#include<set>
using namespace std;
int main(){
  int n, m, i, j=0;
  scanf("%d%d", &n, &m);
  vector<int> pre(m);
  set<int> s;
  for(i=0; i<m; i++){
    scanf("%d", &pre[i]);
    s.insert(pre[i]);
  }
  for(i=0; i<n; i++){
    int u, v, root=-1;
    scanf("%d%d", &u, &v);
    for(j=0; j<m; j++){
      root = pre[j];
      if(root>u&&root<v || root<u&&root>v || root==u || root==v) break;
    }
    if(s.find(u)==s.end()&&s.find(v)==s.end()) printf("ERROR: %d and %d are not found.
", u, v);
    else if(s.find(u)==s.end()) printf("ERROR: %d is not found.
", u);
    else if(s.find(v)==s.end()) printf("ERROR: %d is not found.
", v);
    else if(root==u  || root==v) printf("%d is an ancestor of %d.
", root==u? u : v,  root==u ? v : u);
    else printf("LCA of %d and %d is %d.
", u, v, root);
  }
  return 0;
}