PAT 1143 Lowest Common Ancestor

1143 Lowest Common Ancestor

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.

总结：这是道题目其实挺简单的，主要考察的是对耗时的控制，所以点在于控制好时间复杂度

唉，我只会常规的方法（先建立一颗二叉搜索数，然后遍历二叉搜索树，依据一定的方法找到对一个的位置即可）

代码：

#include 
#include 
using namespace std;
 
struct node{
    int val;
    struct node *left,*right;
};
node *build(node *root,int v){
    if(root==NULL){
        root=new node();
        root->val=v;
        root->left=root->right=NULL;
    }
    else if(v<=root->val)   root->left=build(root->left,v);
    else root->right=build(root->right,v);
    return root;
}
void dfs(node *root,int a,int b){
    if((a>root->val && bval) || (aval && b>root->val)){
        printf("LCA of %d and %d is %d.\n",a,b,root->val);
        return;
    }
    else if(a==root->val){
        printf("%d is an ancestor of %d.\n",a,b);
        return;
    }
    else if(b==root->val){
        printf("%d is an ancestor of %d.\n",b,a);
        return;
    }
    int t=max(a,b);
    if(root->val>t) dfs(root->left,a,b);
    if(root->valdfs(root->right,a,b);
}
int main(){
    int m,n;
    scanf("%d%d",&m,&n);
    int a[n];
    unordered_map<int,int> p;
    
    node *root=NULL;
    for(int i=0;i
        scanf("%d",&a[i]);
        p[a[i]]=1;
        root=build(root,a[i]);
    }
    for(int i=0;i
        int a,b;
        scanf("%d%d",&a,&b);
        if(!p[a] && !p[b])  printf("ERROR: %d and %d are not found.\n",a,b);
        else if(!p[a])  printf("ERROR: %d is not found.\n",a);
        else if(!p[b])  printf("ERROR: %d is not found.\n",b);
        else dfs(root,a,b);
    }
    
    return 0;
}

柳神代码：

简洁明了，想法太妙了！

#include 
#include 
using namespace std;
 
int main(){
    int m,n;
    scanf("%d%d",&m,&n);
    int w[n];
    
    unordered_map<int,bool> p;
    for(int i=0;i
        scanf("%d",&w[i]);
        p[w[i]]=true;
    }
    for(int i=0;i
        int a,b,t;
        scanf("%d%d",&a,&b);
        for(int j=0;j
            t=w[j];
            if((t>a && tb && tbreak;
        }
        if(!p[a] && !p[b])    printf("ERROR: %d and %d are not found.\n",a,b);
        else if(!p[a] || !p[b]) printf("ERROR: %d is not found.\n",!p[a]?a:b);
        else if(t==a || t==b)   printf("%d is an ancestor of %d.\n",t==a?a:b,t==a?b:a);
        else printf("LCA of %d and %d is %d.\n",a,b,t);
    }
    
    return 0;
}

2022.10.31  第一遍已经算是刷完了（接下来好好准备第二遍，明年3月份的考试！）

好好学习，天天向上！

我要考研！