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);
}
}