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.
分析：
法1：遍历输入的序列，找到第一个位于两个数之间的数（BST的性质）
代码：

#include <iostream>
#include <vector>
#include <map>
using namespace std;
map<int, bool> mp;
int main() {
    int m, n, u, v, a;
    scanf("%d %d", &m, &n);
    vector<int> pre(n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &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++) {
            a = pre[j];
            if ((a >= u && a <= v) || (a >= v && a <= u)) break;
        } 
        if (mp[u] == false && mp[v] == false)
            printf("ERROR: %d and %d are not found.
", u, v);
        else if (mp[u] == false || mp[v] == false)
            printf("ERROR: %d is not found.
", mp[u] == false ? u : v);
        else if (a == u || a == v)
            printf("%d is an ancestor of %d.
", a, a == u ? v : u);
        else
            printf("LCA of %d and %d is %d.
", u, v, a);
    }
    return 0;
}
//by liuchuo

法2：

由于前序序列排序之后是中序序列，所以可以根据前序和中序递归求LCA

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int pre[maxn], in[maxn];
int prenum[maxn], innum[maxn];
map<int, int> vis;
int n, m;
int a, b;

int searchroot(int l, int r, int ll, int rr)
{
    int rootpos = innum[pre[l]];
    int posa = innum[a], posb = innum[b];
    if ((rootpos - posa) * (rootpos - posb) <= 0)
        return pre[l];
    if (rootpos > posa)
        return searchroot(l + 1, l + rootpos - ll, ll, rootpos - 1);
    else return searchroot(l + rootpos - ll + 1, r, rootpos + 1, rr);
}

int main()
{
    cin >> m >> n;
    for (int i = 1; i <= n; i++)
        scanf("%d", pre + i), vis[pre[i]] = 1, prenum[pre[i]] = i, in[i] = pre[i];
    sort(in + 1, in + 1 + n);
    for (int i = 1; i <= n; i++)
        innum[in[i]] = i;
    while (m--)
    {
        cin >> a >> b;
        if (!vis[a] && !vis[b]) printf("ERROR: %d and %d are not found.
", a, b);
        else if (!vis[a]) printf("ERROR: %d is not found.
", a);
        else if (!vis[b]) printf("ERROR: %d is not found.
", b);
        else
        {
            int res = searchroot(1, n, 1, n);
            if (res == a)
                printf("%d is an ancestor of %d.
", a, b);
            else if (res == b)
                printf("%d is an ancestor of %d.
", b, a);
            else printf("LCA of %d and %d is %d.
", a, b, res);
        }
    }
}