Lowest Common Ancestor

最新推荐文章于 2021-03-16 16:13:53 发布

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本文介绍了一种在二叉搜索树中寻找两个节点的最低公共祖先(LCA)的方法，利用先序遍历特性，当左子树为最大堆，右子树为最小堆时，大小位于查询节点之间的数即为LCA。文章提供了具体输入输出规范及样例，展示了如何通过遍历树结构来找到LCA，并处理节点不存在的情况。

二叉搜索树，左子树是最大堆，右子树是最小堆，所以根据先序遍历的特点，大小在查询数字之间的数字就是LCA
PAT里关于树的题全部不需要建树 -_-b
第一遍不知道为什么错了，重写一遍就过了

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.

#include<bits/stdc++.h>
using namespace std;
map<int ,int >mp;
int main()
{
	int n,m;
	cin>>n>>m;
	int tree[100005];
	for(int i=0;i<m;i++)
	{
		cin>>tree[i];
		mp[tree[i]]=1;
	}
	for(int i=0;i<n;i++)
	{
		int a,b;
		cin>>a>>b;
		int t;
		for(int j=0;j<m;j++)
		{
			t=tree[j];
			if((t>=a&&t<=b)||(t>=b&&t<=a))break;
		}
		
		if(mp[a]==0&&mp[b]==0)
		printf("ERROR: %d and %d are not found.\n",a,b);
		else if(mp[a]==0||mp[b]==0)
		printf("ERROR: %d is not found.\n",mp[a]==0 ? a:b);
		else if(t==a||t==b)
		printf("%d is an ancestor of %d.\n",t,t==a ? b : a);
		else printf("LCA of %d and %d is %d.\n",a,b,t);
		
	}
}