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 ( $\leq 1,000), the number of pairs of nodes to be tested; and N (\leq 10,000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.$

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

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　　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!!!!!

　　~~~~~~~~~~~

 1 #include <iostream>
 2 #include <vector>
 3 #include <unordered_map>
 4 using namespace std;
 5 int n, m;
 6 vector<int>pre;
 7 unordered_map<int, bool>map;
 8 int main()
 9 {
10     cin >> m >> n;
11     pre.resize(n);
12     for (int i = 0; i < n; ++i)
13     {
14         cin >> pre[i];
15         map[pre[i]] = true;
16     }
17     while (m--)
18     {
19         int a, b;
20         cin >> a >> b;
21         if (map[a] != true && map[b] != true)
22             printf("ERROR: %d and %d are not found.\n", a, b);
23         else if (map[a] != true)
24             printf("ERROR: %d is not found.\n", a);
25         else if (map[b] != true)
26             printf("ERROR: %d is not found.\n", b);
27         else
28         {
29             int k = 0;
30             for (k = 0; k < n; ++k)
31                 if (a <= pre[k] && pre[k] <= b || b <= pre[k] && pre[k] <= a)
32                     break;
33             if (pre[k] == a)
34                 printf("%d is an ancestor of %d.\n", a, b);
35             else if (pre[k] == b)
36                 printf("%d is an ancestor of %d.\n", b, a);
37             else
38                 printf("LCA of %d and %d is %d.\n", a, b, pre[k]);
39         }
40     }
41     return 0;
42 }