PAT甲级——A1143 LowestCommonAncestor【30】

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.

Solution：

　　这道题给出一个重大的提示，就是SBT，题目说明是SBT不是让你自己去兴奋的去重建这棵树【我当时就是这么想的，也这样做了】，而是让你从中发现根节点与左右孩子节点的大小关系，然后从中找到突破口

　　我开始是重建了二叉树，然后DFS来找到两个节点的最低公共节点，然而。。。。。超时了

　　聪明的做法就是从前序遍历中找到突破口【我当时想了，但没有找到规律】

　　首先，使用map来记录哪些节点是存在的，用来判断不存在的节点

　　然后，遍历前序数组，当节点a ，与查询节点 u,v存在关系：（U<=a && a>=v）||(v<=a && u>=a), 那么u，v的最低公共节点就是a!!!!!

　　~~~~~~~~~~~

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
int n, m;
vector<int>pre;
unordered_map<int, bool>map;
int main()
{
    cin >> m >> n;
    pre.resize(n);
    for (int i = 0; i < n; ++i)
    {
        cin >> pre[i];
        map[pre[i]] = true;
    }
    while (m--)
    {
        int a, b;
        cin >> a >> b;
        if (map[a] != true && map[b] != true)
            printf("ERROR: %d and %d are not found.
", a, b);
        else if (map[a] != true)
            printf("ERROR: %d is not found.
", a);
        else if (map[b] != true)
            printf("ERROR: %d is not found.
", b);
        else
        {
            int k = 0;
            for (k = 0; k < n; ++k)
                if (a <= pre[k] && pre[k] <= b || b <= pre[k] && pre[k] <= a)
                    break;
            if (pre[k] == a)
                printf("%d is an ancestor of %d.
", a, b);
            else if (pre[k] == b)
                printf("%d is an ancestor of %d.
", b, a);
            else
                printf("LCA of %d and %d is %d.
", a, b, pre[k]);
        }
    }
    return 0;
}