A1143. 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.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<map>
using namespace std;
typedef struct NODE{
    struct NODE* lchild, *rchild;
    int data;
}node;
int M, N;
int pre[], in[];
map<int, int>mp;
node* create(int preL, int preR, int inL, int inR){
    if(preL > preR)
        return NULL;
    node* root = new node;
    root->data = pre[preL];
    int mid;
    for(int i = inL; i <= inR; i++){
        if(in[i] == root->data){
            mid = i;
            break;
        }
    }
    int len = mid - inL;
    root->lchild = create(preL + , preL + len, inL, mid - );
    root->rchild = create(preL + len + , preR, mid + , inR);
    return root;
}
node* find(node* root, int u, int v){
    if(root == NULL || root->data == u || root->data == v)
        return root;
    node* ll = find(root->lchild, u, v);
    node* rr = find(root->rchild, u, v);
    if(ll != NULL && rr != NULL){
        return root;
    }
    if(ll != NULL){
        return ll;
    }
    if(rr != NULL){
        return rr;
    }
}
int main(){
    scanf("%d%d", &M, &N);
    for(int i = ; i < N; i++){
        scanf("%d", &in[i]);
        pre[i] = in[i];
        mp[pre[i]] = ;
    }
    sort(in, in + N);
    node* root = create(, N - , , N - );
    for(int i = ; i < M; i++){
        int u, v;
        scanf("%d%d", &u, &v);
        if(mp.count(u) ==  && mp.count(v) == ){
            printf("ERROR: %d and %d are not found.\n", u, v);
        }else if(mp.count(u) == ){
            printf("ERROR: %d is not found.\n",u );
        }else if(mp.count(v) == ){
            printf("ERROR: %d is not found.\n",v);
        }else{
            node* ans = find(root, u, v);
            if(ans->data != u && ans->data != v){
                printf("LCA of %d and %d is %d.\n", u, v, ans->data);
            }else if(ans->data == u){
                printf("%d is an ancestor of %d.\n", u, v);
            }else{
                printf("%d is an ancestor of %d.\n", v, u);
            }
        }
    }
    cin >> N;
    return ;
}

总结：

1、题意：给出一个BST的先序序列，再给出两个点u、v，要求在BST中找出uv的最低公共祖先。

2、BST已知先序建树有两种方法，1）先序序列的顺序就是插入顺序，直接依次插入。2）对先序进行排序得到中序序列（BST的中序是从小到大的有序序列），由先序和中序进行递归建树。由于本题的N个数很大，使用insert方法会超时，尤其是在树高度为N时，复杂度为O(n^2)。所以最好采用先序+中序建树。

3、找最低的公共祖先。这种类型的任务一般采用后序递归遍历的办法：先处理左子树，再处理右子树，等左右子树都完成后，综合左右子树返回的信息与root的信息进行某些处理，再返回本层递归的结果。具体到本题，uv只有两种情况：1）即uv分别在某w节点的左右子树，则w为所求。2）uv本身就有祖先后代关系，则若u为祖先，u即为所求。

后序递归，若root为NULL或uv时，说明查找失败或成功，直接返回root。否则说明root为普通节点，先对root的左右子树分别查找。若左右子树都不空时，说明uv分别在root的左右两侧子树，则root即为所求。否则，说明uv在root的一侧子树，若在root的左侧，则将root左侧的查找结果返回。