PAT甲级--1143 Lowest Common Ancestor（30 分）【二叉搜索树最小的父亲结点】

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.

解题思路： 用map来给出现的点做标记，是树上的的就标记为true，其他的默认为false，然后根据前序，来找最小的父亲结点，如果大于等于u且小于等于v或者大于等于v且小于等于u，那就是最小的父亲结点。先判断u和v存不存在，如果都存在再输出最小的父亲结点，这样子少用了变量flag讨论起来也方便。

#include<bits/stdc++.h>
using namespace std;
map<int,bool>mapp;
int main(void)
{
	int n,m;
	scanf("%d %d",&n,&m);
	vector<int>pre(m);
	for(int i=0;i<m;i++)
	{
		scanf("%d",&pre[i]);
		mapp[pre[i]]=true;
	}
	int a;
    for(int i=0;i<n;i++)
    {
    	int u,v;
    	scanf("%d %d",&u,&v);
    	for(int j=0;j<m;j++)
    	{
    		a=pre[j];
    		if(a>=u&&a<=v||a>=v&&a<=u)	break;
	}
		if(mapp[u]==false&&mapp[v]==false) 
		printf("ERROR: %d and %d are not found.\n",u,v);
		else if(mapp[u]==false&&mapp[v]==true)
		printf("ERROR: %d is not found.\n",u);
		else if(mapp[u]==true&&mapp[v]==false)
		printf("ERROR: %d is not found.\n",v);
		else if(a==u||a==v)
		printf("%d is an ancestor of %d.\n",a,a==u?v:u);
		else 
		printf("LCA of %d and %d is %d.\n", u, v, a);
	}
	return 0;
}