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LC - E - 501. Find Mode in Binary Search Tree

Solutions in C++

Solution 1

Complexity Analysis:

• T = O(n log n)

  • inorder(root) - O(n log n), where 'n' = number of nodes

    • Visits every node exactly once.

    • For each node, it does a map[val]++ → this is a logarithmic operation due to std::map.

  • for (auto &elem : m) to find max frequency - O(n)

    • There are at most 'n' unique values.

  • for (auto &elem : m) to collect modes - O(n)

    • There are at most 'n' unique values.

• S = O(n)

  • Map - O(n)

  • Output vector - O(n)

  • Call stack - Recursive in-order traversal

    • For a balanced tree: O(log n), for skewed tree: O(n)

Solution:

x
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/**
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 * Definition for a binary tree node.
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 * struct TreeNode {
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 *     int val;
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 *     TreeNode *left;
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 *     TreeNode *right;
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 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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 * };
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 */
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class Solution {
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public:
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    vector<int> findMode(TreeNode* root) {
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        vector<int> v;
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        if (root == nullptr)
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        {
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            v.push_back(0);
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            return v;
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        }
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        // populate the map
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        inorder(root);
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        // find the maximum occurrence
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        int max = 0;
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        for (auto &elem : m)
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        {
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            if (elem.second > max)
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                max = elem.second;
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        }
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        // populate the vector with the key(s) with maximum occurrence
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        for (auto &elem : m)
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        {
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            if (elem.second == max)
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                v.push_back(elem.first);
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        }
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        return v;
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    }
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private:
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    void inorder(TreeNode *root)
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    {
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        // recursively traverse the passed tree and populate the map
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        // terminating condition
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        if (root == nullptr)
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            return;
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        // recursive steps
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        inorder(root->left);
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        m[root->val]++;
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        inorder(root->right);
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        return;
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    }
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    std::map<int, int> m;
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};