501. Find Mode in Binary Search Tree

• Difficulty: Easy

• Topics: Tree

• Similar Questions:

  • Validate Binary Search Tree

Problem:

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to 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.

For example:
Given BST [1,null,2,2],

   1
    \
     2
    /
   2

return [2].

Note: If a tree has more than one mode, you can return them in any order.

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

Solutions:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> findMode(TreeNode* root) {
        if (root == nullptr)    return {};
        int count = 0;
        vector<int> ret;
        int lastVal = 0;
        int largest = 0;

        helper(root, lastVal, count, ret, largest);
        if (count > largest) {
            ret.clear();
            ret.push_back(lastVal);
        } else if (count == largest) {
            ret.push_back(lastVal);
        }

        return ret;
    }

private:
    void helper(TreeNode* root, int& lastVal, int& count, vector<int>& ret, int& largest) {
        if (root == nullptr)    return;
        helper(root->left, lastVal, count, ret, largest);

        if (root->val == lastVal) {
            ++count;
        } else {
            if (count > 0) {
                if (count > largest) {
                    ret.clear();
                    ret.push_back(lastVal);
                    largest = count;
                } else if (count == largest) {
                    ret.push_back(lastVal);
                }
            }
            count = 1;
        }

        lastVal = root->val;

        helper(root->right, lastVal, count, ret, largest);
    }

};