1143 Lowest Common Ancestor （30 分）

寻找二叉搜索树中两节点的最近公共祖先

最新推荐文章于 2020-04-28 16:17:53 发布

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本文介绍了一种算法，用于在给定的二叉搜索树中找到两个节点的最低公共祖先（LCA）。通过分析二叉搜索树的性质，即左子树的所有节点小于根节点，右子树的所有节点大于等于根节点，可以有效地找到LCA。输入包括节点对的数目、树的大小和先序遍历序列，输出则是每对节点的LCA或错误信息。

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.

本题给的二叉树是二叉排序树的先序序列，如果uv在根的左右，那么此点符合要求。注意要从是第一个符合要求的点，先序：根左右

#include<iostream>
#include<map>
#include<vector>
#include<cstdio>
using namespace std;
int main(){
	int m,n;
	map<int,bool> mp;
	cin >> m >> n;
	vector<int> v(n+1);
	for(int i =0;i < n;i++){
		cin >> v[i];
		mp[v[i]] = true;
	}
	while(m--){
		int a,b;
		cin >> a >> b;
		bool flag = false;
		int ans;
		for(int i = 0;i < n;i++){
			
			if(((v[i]>=a&&v[i]<=b)||(v[i]>=b&&v[i]<=a))&&mp[a]&&mp[b]){
				flag = true;
				ans = v[i];
				break;
			}
		}
		if(flag){
			if(ans==a){
				printf("%d is an ancestor of %d.\n",a,b);
			}else if(ans==b){
				printf("%d is an ancestor of %d.\n",b,a);
			}else{
				printf("LCA of %d and %d is %d.\n",a,b,ans);
			}
		}else{
			if(mp[a]==false&&mp[b]==false){
				printf("ERROR: %d and %d are not found.\n",a,b);
			}else if(mp[a]==false){
				printf("ERROR: %d is not found.\n",a);
			}else if(mp[b]==false){
				printf("ERROR: %d is not found.\n",b);
			}
		}
	}
	return 0;
}