Question

给定二进制搜索树，其中可能包含重复项，但BST的所有其他逻辑都是完整的，确定最常出现的元素。

class TreeNode
{
public:
    TreeNode* right = NULL;
    TreeNode* left = NULL;
    int val;

    TreeNode(int value)
    {
        val = value;
    }
};

// To keep track of the frequency of the value/node
struct holder
{
public:
    TreeNode* most = NULL;
    int count = 0;
};

int frequencyOfNode(TreeNode* root, struct holder* ptr)
{
    if (root == NULL)
    {
        return 0;
    }

    int left = frequencyOfNode(root->left, ptr);
    int right = frequencyOfNode(root->right, ptr);

// need to check of left and right are nor null
    if (left != 0 && root->val == root->left->val)
    {
        return 1 + left;
    }
    else if (right != 0 && root->val == root->right->val)
    {
        return 1 + right;
    }
    else
    {
        // left has a higher frequency
        if (left >= right)
        {
            // left is bigger;
            if (left > ptr->count)
            {
                ptr->most = root->left;
                ptr->count = left;
            }
        }
        else
        {
            // right has a higher frequency
            if (right > ptr->count)
            {
                ptr->most = root->right;
                ptr->count = right;
            }
        }

        return 1;
    }
}

我正在对二叉搜索树进行后期遍历。当节点按连续顺序出现时，我的逻辑工作，但如果节点不是连续的顺序;重置节点的频率。

我的时间是O（n），空格是O（1）。

问题是当节点没有连续链接时。

我的示例树：

int main()
{
    TreeNode *root = new TreeNode(6);
    root->right = new TreeNode(8);
    root->right->left = new TreeNode(7);
    root->right->right = new TreeNode(8);
    root->right->right->right = new TreeNode(8);
    root->right->right->right->right = new TreeNode(9);
    root->right->right->right->right->left = new TreeNode(8);
    root->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->left->right->right = new TreeNode(5);
    root->left->right->right->right = new TreeNode(5);
    root->left->left = new TreeNode(1);
    root->left->left->right = new TreeNode(1);
    root->left->left->right->right = new TreeNode(1);
    root->left->left->right->right = new TreeNode(2);
    root->left->left->left = new TreeNode(0);

    struct holder freq;
    int ran = frequencyOfNode(root, &freq);
    std::cout << "random" << ran << std::endl;

    std::cout << "The Node: " << freq.most->val << " frequency " << freq.count
            << std::endl;

    return 0;
}

当节点不连续时（即8-> 8-> 8-> 9-> 8），我真的很困惑如何考虑。

Answer 1

我看到你自己解决了一些问题。无论如何，我决定完全解决这个问题，改变一些事情以简化一切。它使用O（N）时间和O（1）空间：

#include <iostream>
#include <limits>

class TreeNode
{
public:
    TreeNode* right;
    TreeNode* left;
    int val;

    TreeNode(int value)
    {
        val = value;
        right = left = NULL;
    }
};

// To keep track of the frequency of the value/node
struct Holder
{
public:
    int value;
    int count;

    Holder(int v=std::numeric_limits<int>::min(), int c=-1): value(v), count(c) {}
};



void dfs(TreeNode* root, int &mostFrequent, int &mostFrequentCount, int &current, int &currentCount)
{
    if(root->left) dfs(root->left, mostFrequent, mostFrequentCount, current, currentCount); //first go to smaller

    int val = root->val;

    if(val == current) currentCount++;
    else { current=val; currentCount=1; }

    if(currentCount > mostFrequentCount) 
    {
        mostFrequent=current;
        mostFrequentCount=currentCount;
    }

    if(root->right) dfs(root->right, mostFrequent, mostFrequentCount, current, currentCount); //finally go to larger
}

Holder getMostFrequent(TreeNode *root)
{
    int mostFrequent=-1,mostFrequentCount=-1, current=std::numeric_limits<int>::min(), currentCount=-1;
    if(root) dfs(root, mostFrequent, mostFrequentCount, current, currentCount);

    return Holder(mostFrequent, mostFrequentCount);
}


int main()
{
    TreeNode *root = new TreeNode(6);
    root->right = new TreeNode(8);
    root->right->left = new TreeNode(7);
    root->right->right = new TreeNode(8);
    root->right->right->right = new TreeNode(8);
    root->right->right->right->right = new TreeNode(9);
    root->right->right->right->right->left = new TreeNode(8);
    root->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->left->right->right = new TreeNode(5);
    root->left->right->right->right = new TreeNode(5);
    root->left->left = new TreeNode(1);
    root->left->left->right = new TreeNode(1);
    root->left->left->right->right = new TreeNode(1);
    root->left->left->right->right = new TreeNode(2);
    root->left->left->left = new TreeNode(0);

    Holder h = getMostFrequent(root);

    std::cout << "most frequently encountered element: " << h.value << ", " << h.count << " times\n";


    return 0;
}

它使用的事实是，因为这是一个BST，所以在[left - ＆gt;中遍历它]当前 - ＆gt;正确的顺序将导致排序的元素，这就是全部。

二叉搜索树中节点/值的频率

1 个答案: