501. Find Mode in Binary Search Tree

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My Solution

The Key Idea for Solving This Coding Question

C++ Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    vector<int> findMode(TreeNode *root) {
        predecessorVal = -200000;
        repeatCnt = 1;
        maxRepeat = 1;

        dfs(root);
        return answer;
    }

private:
    int predecessorVal;
    int repeatCnt;
    int maxRepeat;
    vector<int> answer;

    void dfs(TreeNode *root) {
        if (root == nullptr) {
            return;
        }

        dfs(root->left);

        if (predecessorVal == root->val) {
            ++repeatCnt;
        } else {
            repeatCnt = 1;
        }

        if (maxRepeat < repeatCnt) {
            maxRepeat = repeatCnt;
            answer.clear();
            answer.push_back(root->val);
        } else if (maxRepeat == repeatCnt) {
            answer.push_back(root->val);
        }

        predecessorVal = root->val;

        dfs(root->right);
    }
};

Time Complexity

$O(n)$

$n$ is the number of nodes in the binary tree referred by root.

Space Complexity

$O(H)$

$H$ is the height of the binary tree referred by root.

Miscellane