LCA_1143 Lowest Common Ancestor （30 分）

目录

1143 Lowest Common Ancestor

 解题思路

程序

参考博客

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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.

 解题思路

给出二叉搜索树的先序遍历，计算给出两点的最近公共祖先

因为给出的是二叉搜索树的先序遍历，首先是先序遍历，任意两点的公共祖先一定出现在他们两点的前面，因为祖先都在前面已经输出了

然后是二叉搜索树，左边子树都小于根节点，右边子树都大于根节点，那么他们的公共祖先，大小一定位于两点之间，即假设两点为x，y，则公共祖先root，一定为root>=x&&root<=y或root>=y&&root<=x

上面所说，我们直接遍历先序数组，然后找这样的root即可

程序

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f

using namespace std;
const int maxn = 1e4+5;
int pre[maxn];
int n,m;
unordered_map pos;

int main()
{
    scanf("%d%d",&n,&m);
    for(int i =1 ;i <= m;i ++)
    {
        scanf("%d",pre+i);
        pos[pre[i]] = i;
    }
    int x,y;
    while(n--)
    {
        scanf("%d%d",&x,&y);
        if(pos[x] == 0 && pos[y] == 0)
            printf("ERROR: %d and %d are not found.\n",x,y);
        else if(pos[x] == 0 || pos[y] == 0)
            printf("ERROR: %d is not found.\n",pos[x]==0?x:y);
        else
        {
            int p = min(pos[x],pos[y]);
            for(int i = 1;i <= p;i ++)
                if((pre[i] >= x && pre[i] <= y) || (pre[i] <= x && pre[i] >= y))
                {
                    if(pre[i] == x || pre[i] == y)
                        printf("%d is an ancestor of %d.\n",pre[i],pre[i]==x?y:x);
                    else
                        printf("LCA of %d and %d is %d.\n",x,y,pre[i]);
                    break;
                }
        }
    }
    return 0;

}

参考博客

https://www.liuchuo.net/archives/4616