PAT Advanced—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.

题目大意：
  l求出两个结点的最近共同祖先结点

思路：
  主要利用二叉查找树的性质，结点的左子树的键都比该结点小，右子树的键都比该结点大，只要两个结点分别在不同的子树上，则该结点就为这两个结点的最小公共祖先结点。

代码：(C++)

#include
#include
#include

using namespace std;

map<int,bool> mp;  //map的bool值初始化为false

int main(){
	int m,n,u,v,a;
	cin>>m>>n;
	vector<int> pre(n);
	for(int i=0; i<n; i++){
		cin>>pre[i];
		mp[pre[i]] = true;
	}
	for(int i=0; i<m; i++){
		scanf("%d %d", &u, &v);
		for(int j=0; j<n; j++){
			/*
                主要先序遍历的二叉搜索树，只要判断结点的key就可以判断
                该结点在根结点的左子树还是右子树
			*/
			a = pre[j];  //从第一个结点开始判断
			if((a>=u && a<=v) || (a<=u && a>=v)){  //结点在两边，则当前结点就是最近祖先结点
				break;
			}
		}
		if(mp[u] == false && mp[v] == false)
			printf("ERROR: %d and %d are not found.\n", u, v);  //bool值用来判断是否存在这个结点
		else if(mp[u] == false || mp[v] == false)
			printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
		else if(a == v || a == u)
			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;
}

参考：
  https://www.liuchuo.net/archives/4616