PAT 甲级1143 Lowest Common Ancestor (30 分,一个测试点未过)

二叉搜索树中寻找最低公共祖先
本文探讨了在二叉搜索树中寻找两个指定节点的最低公共祖先(LCA)的问题,提供了详细的算法解释及代码实现。通过分析先序遍历序列,利用map和set数据结构,有效地解决了LCA查找问题。

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.

这一题是较为简单的lca问题,尽管最后还是有一个测试点超时了(仍然未解决)。这个问题的解决方法便是先手工实现以找到规律,而后采用代码的方法解出。他给出了这个序列是先序序列,而后这个先序序列节点的顺序是左孩子右孩子必须在根节点之后,而且左孩子比根大,右孩子比根小,(这读了好久的题才读懂)。而后,需要做的是找规律,这里找到的规律是其lca首先要在两节点的位置之前,其次,值要在两个节点大小之间。若在两点之前找不到对应的点,则两点是父子关系,出现在前的便是父亲节点。 

因此,这里定了两个map(是不是麻烦了些?)一个是下标对数值,另一个是数值对下标,以解决最前公共子节点问题。第二个注意的地方是设置一个set,用来存放输入的数值。以此来确定节点是不是在树内。

完整代码如下(还有超时的一部分不知道怎么改):

#include <bits/stdc++.h>

using namespace std;

vector <int> pre;

set<int> s;

map<int,int>pos,pos2;

void pre_order(int c,int d)
{
    int pos_c=pos2[c];
    int pos_d=pos2[d];
    int max_p=pos_c<pos_d?pos_c:pos_d;//max_p 用于找出c与d两点哪一个位置在前
    int root=0;
    int flag=0;
    for(int i=1;i<max_p;i++)
    {
        if((pos[i]>c&&pos[i]<d)||(pos[i]<c&&pos[i]>d))
        {
            root=pos[i];
            cout<<"LCA of "<<c<<" and "<<d<<" is "<<root<<"."<<endl;
            flag=1;
            break;
        }
    }
    if(flag==0)
    {
        if(max_p==pos_c)
        {
            cout<<c<<" is an ancestor of "<<d<<"."<<endl;
        }
        else
        {
            cout<<d<<" is an ancestor of "<<c<<"."<<endl;
        }
    }
}
int main()
{
    int n,m;
    cin>>n>>m;
    int i=0;
    pre.resize(m+1);
    for(i=1;i<=m;i++)
    {
        cin>>pre[i];
        pos[i]=pre[i];
        pos2[pre[i]]=i;
        s.insert(pre[i]);
    }
    int c,d;
    for(i=1;i<=n;i++)
    {
        cin>>c>>d;
        if(s.find(c)==s.end()&&s.find(d)==s.end())
        {
            cout<<"ERROR: "<<c<<" and "<<d<<" are not found."<<endl;
        }
        else if(s.find(c)!=s.end()&&s.find(d)==s.end())
        {
            cout<<"ERROR: "<<d<<" is not found."<<endl;
        }
        else if(s.find(c)==s.end()&&s.find(d)!=s.end())
        {
            cout<<"ERROR: "<<c<<" is not found."<<endl;
        }
        else
        {
            pre_order(c,d);
        }
    }
    return 0;
}

 

以下是C#中二叉树的lowest common ancestor的源代码: ```csharp using System; public class Node { public int value; public Node left; public Node right; public Node(int value) { this.value = value; this.left = null; this.right = null; } } public class BinaryTree { public Node root; public BinaryTree() { this.root = null; } public Node LowestCommonAncestor(Node node, int value1, int value2) { if (node == null) { return null; } if (node.value == value1 || node.value == value2) { return node; } Node left = LowestCommonAncestor(node.left, value1, value2); Node right = LowestCommonAncestor(node.right, value1, value2); if (left != null && right != null) { return node; } return (left != null) ? left : right; } } public class Program { public static void Main() { BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); Node lca = tree.LowestCommonAncestor(tree.root, 4, 5); Console.WriteLine("Lowest Common Ancestor of 4 and 5: " + lca.value); lca = tree.LowestCommonAncestor(tree.root, 4, 6); Console.WriteLine("Lowest Common Ancestor of 4 and 6: " + lca.value); lca = tree.LowestCommonAncestor(tree.root, 3, 4); Console.WriteLine("Lowest Common Ancestor of 3 and 4: " + lca.value); lca = tree.LowestCommonAncestor(tree.root, 2, 4); Console.WriteLine("Lowest Common Ancestor of 2 and 4: " + lca.value); } } ``` 在上面的代码中,我们定义了一个Node类和一个BinaryTree类。我们使用BinaryTree类来创建二叉树,并实现了一个LowestCommonAncestor方法来计算二叉树中给定两个节点的最近公共祖先。 在LowestCommonAncestor方法中,我们首先检查给定节点是否为null或与给定值之一匹配。如果是,则返回该节点。否则,我们递归地在左子树和右子树上调用LowestCommonAncestor方法,并检查它们的返回值。如果左子树和右子树的返回值都不为null,则当前节点是它们的最近公共祖先。否则,我们返回非null的那个子树的返回值。 在Main方法中,我们创建了一个二叉树,并测试了LowestCommonAncestor方法的几个不同输入。
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