17年冬季第四题 PAT甲级 1143 Lowest Common Ancestor（30分）

港控/mmm° 2023-07-10 11:17 6阅读 0赞

题目来源：[https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312][https_pintia.cn_problem-sets_994805342720868352_problems_994805343727501312]

备考汇总贴：[2020年3月PAT甲级满分必备刷题技巧][2020_3_PAT]

# 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.

# 题目大意（以下内容主要转载自[liuchuo][]，补充了讲解） #

给出一棵二叉搜索树的前序遍历，问结点u和v的共同最低祖先是谁

# 题目分析 #

1.二叉搜索树的前序遍历有个特点：排在前面的某个节点一定是“在它后面且不大于它的节点”的父亲。所以，将当前结点标记为a，如果u和v分别在a的左、右，或者u、v其中一个就是当前a，说明找到了这个共同最低祖先a，退出当前循环

2.map<int, bool> mp用来标记树中所有出现过的结点，遍历一遍pre数组

# 满分代码 #

#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.\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;
    }

# 其他解法（转载自[其他网友][Link 1]） #

建树、递归查找节点看看符不符合lca的定义。

#include<iostream>
    #include<unordered_map>
    #include<algorithm>
    #include<unordered_set>
    #include<string>
    using namespace std;
    #define MAX 10005
    int pre[MAX],in[MAX],N,M;
    
    typedef struct node{
        int left,right;
    }node;
    unordered_map<int,node>arr;
    void CreatBST(int preL,int inL,int inR)
    {
        if(inL>inR)return ;
        int i=inL;
        while(pre[preL]!=in[i])++i;
        if(i-inL)arr[pre[preL]].left=pre[preL+1];
        else arr[pre[preL]].left=0;
        if(inR-i)arr[pre[preL]].right=pre[preL+i-inL+1];
        else arr[pre[preL]].right=0;
        CreatBST(preL+1,inL,i-1);
        CreatBST(preL+i-inL+1,i+1,inR);
    }
    void findLCA(int v,int a,int b)
    {
        if(a==v)printf("%d is an ancestor of %d.\n",a,b);
        else if(b==v)printf("%d is an ancestor of %d.\n",b,a);
        else if(a<v&&b>v||a>v&&b<v)printf("LCA of %d and %d is %d.\n",a,b,v);
        else if(a<v&&b<v)findLCA(arr[v].left,a,b);
        else findLCA(arr[v].right,a,b);
    }
    int main()
    {
        //freopen("test.txt","r",stdin);
        int i,u,v,root;
        scanf("%d %d",&M,&N);
        for(i=0;i<N;++i){
            scanf("%d",&pre[i]);
            if(i==0)root=pre[0];
            in[i]=pre[i];
        }
        sort(in,in+N);
        CreatBST(0,0,N-1);
        while(M--){
            scanf("%d %d",&u,&v);
            if(!arr.count(u)&&!arr.count(v))printf("ERROR: %d and %d are not found.\n",u,v);
            else if(!arr.count(u)||!arr.count(v))printf("ERROR: %d is not found.\n",arr.count(u)?v:u);
            else findLCA(root,u,v);
        }
        return 0;
    }

[https_pintia.cn_problem-sets_994805342720868352_problems_994805343727501312]: https://pintia.cn/problem-sets/994805342720868352/problems/994805343727501312
[2020_3_PAT]: https://blog.csdn.net/allisonshing/article/details/104213959
[liuchuo]: https://www.liuchuo.net/archives/4616
[Link 1]: https://blog.csdn.net/lianwaiyuwusheng/article/details/87861627