PAT (Advanced Level) Practice 1143 Lowest Common Ancestor (30分) （LCA+二叉搜索树）

1.题目

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.

2.题目分析

1.这道题目是1151 LCA in a Binary Tree (30分)的简化版，我是先做的1151，用了四种方法（https://blog.csdn.net/qq_42325947/article/details/104975490），所以这道题就很简单啦

（https://pintia.cn/problem-sets/994805342720868352/problems/1038430130011897856 ）

2.二叉搜索树的中序遍历是由小到大的，所以本题未给出中序遍历，也不用专门记录中序遍历，使用1151 LCA in a Binary Tree的方法一（方法一的详细解释在上面的第一个链接），是不过pos只用于查询是否存在，在LCA函数中通过数字大小判断左右子树的位置

3.代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<functional>
#include<unordered_map>
using namespace std;
int pre[10002];
unordered_map<int, int>pos;
void lca(int root, int aa, int bb)
{
	if (pre[root]>aa && pre[root] > bb)
		lca(root + 1, aa, bb);
	else if ((pre[root]<aa && pre[root] > bb) || (pre[root]>aa && pre[root] < bb))
		printf("LCA of %d and %d is %d.
", aa, bb, pre[root]);
	else if (pos[pre[root]] < pos[aa] && pos[pre[root]] < pos[bb])
	{
		int temp = root;
		while (pre[temp] <= pre[root])temp++;
		lca(temp,aa, bb);
	}
	else if (aa ==pre[root])
		printf("%d is an ancestor of %d.
", aa, bb);
	else if (bb== pre[root])
		printf("%d is an ancestor of %d.
", bb, aa);
}

int main()
{
	int m, n;
	scanf("%d %d", &m, &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &pre[i]);
		pos[pre[i]] = i;
	}
	for (int i = 1; i <= m; i++)
	{
		int aa, bb;
		scanf("%d %d", &aa, &bb);
		if (pos[aa] == 0 && pos[bb] != 0)
			printf("ERROR: %d is not found.
", aa);
		else 	if (pos[aa] != 0 && pos[bb] == 0)
			printf("ERROR: %d is not found.
", bb);
		else	if (pos[aa] == 0 && pos[bb] == 0)
			printf("ERROR: %d and %d are not found.
", aa, bb);
		else 	if (pos[aa] != 0 && pos[bb] != 0)
			lca(1, aa, bb);
	}
}