1143 Lowest Common Ancestor (30 分)

2019-05-13

1143 Lowest Common Ancestor (30 分)

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.

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一棵二叉树，求给定两个结点的最近公共祖先。

分析

一开始想用记录父结点的办法，但是结点值会有重复，无法实现。后来在网上看到，因为已经给出了先序遍历，只要按照先序遍历依次判断。假设当前结点为a，如果u,v都小于a，那么最近公共祖先必在a左子树上。同理，大于a则在右子树。如果a正好大于等于其中一个小于等于另一个，那么a则为最近公共祖先。

代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>

using namespace std;
int m,n,head;
map<int,int> showup;
vector<int> pre;

int main() {
    cin>>m>>n;
    int t;
    for(int i=0;i<n;i++){
        cin>>t;
        showup[t]=1;
        pre.push_back(t);
    }
    int u,v;
    for(int i=0;i<m;i++){
        cin>>u>>v;
        int a;
        for(int j=0;j<pre.size();j++){
            a=pre[j];
            if((a>=u&&a<=v)||(a>=v&&a<=u))
                break;
        }
        if(showup[u]==0&&showup[v]==0)
            printf("ERROR: %d and %d are not found.\n", u, v);
        else if(showup[u]==0)
            printf("ERROR: %d is not found.\n", u);
        else if(showup[v]==0)
            printf("ERROR: %d is not found.\n", 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;
}