题目描述

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 binary tree, 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.

考点

1.二叉搜索树、先序遍历、最小公共父结点。

思路

递归实现
判断根结点与两个结点的相对位置关系，如果两个结点都在根结点的左边，那么它们的最小公共父结点一定在根节点的左子树上；在右边的情况类似。如果分别位居左右，那么这个根节点就是两个结点的最小父结点。

代码

递归实现

#include <iostream>
#include <vector>
#include <map>
using namespace std;
vector<int> pre;
map<int, int>pos;
void LCA(int a, int b, int r) {
    if (a < pre[r] && b < pre[r]) {
        LCA(a, b, r + 1);
    }
    else if ((a < pre[r] && b > pre[r]) || (a > pre[r] && b < pre[r])) {
        cout << "LCA of " << a << " and " << b << " is " << pre[r] << "." <<endl;
    }
    else if (a > pre[r] && b > pre[r]) {
        LCA(a, b, r + 1);
    }
    else if (a == pre[r]) {
        cout << a << " is an ancestor of " << b << "." << endl;
    }
    else if (b == pre[r]) {
        cout << b << " is an ancestor of " << a << "." << endl;
    }
}
int main() {
    int m, n;
    cin >> m;
    cin >> n;
    pre.resize(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> pre[i];
        pos[pre[i]] = i;
    }
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a;
        cin >> b;
        if (pos[a] == 0 && pos[b] == 0) {
            cout << "ERROR: " << a << " and " << b << " are not found." << endl;
        }
        else if (pos[a] == 0 || pos[b] == 0) {
            cout << "ERROR: " << (pos[a] == 0 ? a : b) << " is not found." << endl;
        }
        else {
            LCA(a, b, 1);
        }
    }
    return 0;
}

贴一个柳神的代码，循环实现

#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.\n", u, v);
        else if (mp[u] == false || mp[v] == false)
            printf("ERROR: %d is not found.\n", mp[u] == false ? u : v);
        else if (a == u || a == v)
            printf("%d is an ancestor of %d.\n", a, a == u ? v : u);
        else
            printf("LCA of %d and %d is %d.\n", u, v, a);
    }
    return 0;
}

果然柳神的代码看起来简单多了！

1143.Lowest Common Ancestor