1143 Lowest Common Ancestor

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.

分析

本题是将BST和LCA算法结合起来进行考察，看起来很复杂，实际上是降低了难度。最直接的想法就是先利用pre数组构建出一个BST，然后在用LCA算法搜索结果，但是题目限制了时间为200ms，这样的话部分样例无法通过。题解的思路就非常巧妙，直接对pre数组进行遍历，每次比较带查询数据x，y和当前元素的大小关系即可确定位置。原理是利用了BST的属性，大小关系即确定了元素之间的位置关系。大的元素一点在右，小的元素一定在左，因此只要满足对于一个元素a，x>=a&&y<=a或者y>=a&&x<=a即可说明x，y在a的两侧，即公共节点。其他边界条件再加以判断即可。另题解中有另外一个技巧：利用一个map来存储元素是否存在，提高了速度。

代码

//
// Created by masterCai on 2020/5/14.
//

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

int main(){
    int M, N;
    cin >> M >> N;
    vector<int> pre(N);
    unordered_map<int, bool> m;
    for (int i = 0; i < N; ++i) {
        scanf("%d", &pre[i]);
        m[pre[i]] = true;
    }
    for (int i = 0; i < M; ++i) {
        int x, y;
        scanf("%d %d", &x, &y);
        int a;
        for (int j = 0; j < N; ++j) {
            a = pre[j];
            if((a>=x && a <= y)||(a>=y && a<=x)){
                break;
            }
        }
        if(!m[x] && !m[y]){
            printf("ERROR: %d and %d are not found.\n", x, y);
        } else if(!m[x] || !m[y]){
            printf("ERROR: %d is not found.\n", !m[x] ? x : y);
        } else if(a==x || a== y){
            printf("%d is an ancestor of %d.\n", a, a == x ? y : x);
        } else{
            printf("LCA of %d and %d is %d.\n", x, y, a);
        }
    }
    return 0;
}