leetcode-501. Find Mode in Binary Search Tree

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).

 

思路：

1.递归记录每个节点，同时返回最大的相同节点数。

2.使用map 哈希思想是解决这种题目的最好选择

 

Accepted Code:

/**
 * 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) {
        unordered_map<int,int> mp;
        vector<int> result;
        int modeCount=getModeCount(root,mp);
        
        for(pair<int,int> p:mp)
        {
            if(p.second==modeCount)
            result.push_back(p.first);
        }
        return result;
    }
    
    int getModeCount(TreeNode* root,unordered_map<int,int>& mp)
    {
        if(root==nullptr)
        return 0;
        if(mp.find(root->val)==mp.end())
        {
            mp.insert(pair<int,int>(root->val,1));
        }else
        {
            mp[root->val]++;
        }
        return max(mp[root->val],max(getModeCount(root->left,mp),getModeCount(root->right,mp)));
    }
};