Question

我试图在我的二叉搜索树中找到多数元素，但我没有运气。任何想法如何实现获取多数元素的算法？我认为我的二叉搜索树没问题，但我真的不知道如何使用二叉搜索树搜索多数元素这是我的代码

class BinarySearchTree
{
    private Node root = null;
    public Node Root { get => root; set => root = value; }

    private int size = 0;
    public BinarySearchTree()
    {
        //empty
    }


    public Node getNodeByValue(int v, Node start)
    {
        //if start node is empty value not found.
        if (start == null)
        {
            return null;
        }

        if (v == start.Value)
        {
            return start; //Value found at start
        }
        else if (v < start.Value)
        {

            return getNodeByValue(v, start.LeftChild);
        }

        else if (v > start.Value)
        {
            return getNodeByValue(v, start.RightChild);
        }
        else
        {
            return null; //value not found  
        }
    }


    public void AddNode(Node start, int v)
    {
        //Insert the new node in the tree
        InsertNewNode(v, start);
    }


    public Node InsertNewNode(int v, Node start)
    {
        Node newNode = new Node(v);
        if (root == null)
        {
            root = newNode;
        }

        if (start == null)
        {
            start = newNode;
            size++;
            return start;
        }
        if (v < start.Value)
        {

            if (start.LeftChild == null)
            {

                start.LeftChild = newNode;
            }
            else
            {
                InsertNewNode(v, start.LeftChild);
            }
        }
        else if (v > start.Value)
        {
            if (start.RightChild == null)
            {

                start.RightChild = newNode;
            }
            else
            {
                InsertNewNode(v, start.RightChild);
            }
        }
        else
        {
            InsertNewNode(v, start);
            size++;
        }
        return start;
    }


    public void IncrementNodeFrequency(Node n)
    {
        //If the node is not null
        if (n != null)
        {

            n.Frequency = n.Frequency + 1;
        }
    }


    public int getMajorityElement(Node start)
    {

        if(start!=null)
        {

        }


    }

}

提前谢谢！

Answer 1

在许多访谈中都会提出这个问题，它有很多方法：

1）遍历整个树并包含在字典中并获得最多计数

2）或者在职业杯中的这个问题：https://www.careercup.com/question?id=17192662

#include <iostream>
using namespace std;

struct TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int val_)
    :val(val_), left(NULL), right(NULL){}
};

int maxFreqVal;
int maxFreqCount;

void inorder(TreeNode *node, int &curFreqVal, int &curFreqCount){
    if(node == NULL)
    return;

    inorder(node->left, curFreqVal, curFreqCount);
    if(curFreqVal != node->val){
    curFreqVal = node->val;
    curFreqCount = 1;
    }else{
    curFreqCount++;
    if(curFreqCount > maxFreqCount){
        maxFreqVal = curFreqVal;
        maxFreqCount = curFreqCount;
    }
    }
    inorder(node->right, curFreqVal, curFreqCount);
}

int getFreq(TreeNode *root){
    if(root == NULL)
    return -1;

    maxFreqVal = root->val;
    maxFreqCount = 1;
    int curFreqVal = root->val;
    int curFreqCount = 1;
    inorder(root, curFreqVal, curFreqCount);
    return maxFreqVal;
}

int main(){
    TreeNode *root = new TreeNode(10);
    root->right = new TreeNode(12);
    root->right->left = new TreeNode(12);
    root->right->left->left = new TreeNode(12);
    root->right->left->right = new TreeNode(12);
    root->right->right = new TreeNode(12);
    root->right->right->left = new TreeNode(12);
    root->right->right->right = new TreeNode(12);
    root->left = new TreeNode(6);
    root->left->left = new TreeNode(6);
    root->left->left->left = new TreeNode(4);
    root->left->left->right = new TreeNode(6);
    root->left->right = new TreeNode(6);
    root->left->right->left = new TreeNode(6);
    root->left->right->right = new TreeNode(6);
    root->left->right->right->right = new TreeNode(7);
    int result = getFreq(root);
    cout << "result : " << result << ", frequency : " << maxFreqCount << endl;
    return 0;
}

使用二叉搜索树获取多数元素c＃

1 个答案: