【LeetCode 42】501. Mode in binary search tree

【LeetCode 42】501. The mode in the binary search tree

One 、 The question

Two 、 The answer process

stay 《 Binary tree ： The minimum absolute value difference of the search tree 》 We used pre Pointers and cur Skills of pointer . Here again .

Make a pointer to the previous node , So every time cur（ Current node ） Talent and pre（ Previous node ） The comparison .

And when initializing pre==NULL, So when pre by NULL when , We know that this is the first element of comparison .

if (pre == NULL) {
     //  First node 
    count = 1; //  The frequency is 1
} else if (pre->val == cur->val) {
     //  Same value as the previous node 
    count++;
} else {
     //  Different from the previous node value 
    count = 1;
}
pre = cur; //  Update previous node 

How to traverse once to find all modes ？

Traversal frequency count Greater than maxCount When , Not only to update maxCount, And empty the result set （ The following code is result Array ）, Because the elements before the result set are invalid .

if (count > maxCount) {
     //  If the count is greater than the maximum 
    maxCount = count;   //  Update maximum frequency 
    result.clear();     //  A crucial step , Don't forget to empty result, Before result All the elements in the have failed 
    result.push_back(cur->val);
}

class Solution {
    
public:
    int maxCount;// Maximum frequency 
    int count;// Statistical frequency 
    TreeNode* pre;
    vector<int>result;
    void searchBST(TreeNode *cur)
    {
    
        if(cur==NULL) return;
        searchBST(cur->left);// Left 

        if(pre==NULL)// The first node 
        {
    
            count=1;
        }else if(pre->val==cur->val){
    // Same value as the previous node 
            count++;
        } else{
    // Different from the previous node value 
            count=1;
        }
        pre=cur;// Update previous node 

        if(count==maxCount)// If it is the same as the maximum , In the result
        {
    
            result.push_back(cur->val);
        }
        if(count>maxCount)// If the count is greater than the maximum frequency 
        {
    
            maxCount=count;// Update maximum frequency 
            result.clear();// Key step , Don't forget to empty result, Before result All the elements in the have failed 
            result.push_back(cur->val);
        }

        searchBST(cur->right);// Right 
        return;
    }

    vector<int> findMode(TreeNode* root) {
    
        count=0;
        maxCount=0;
        TreeNode* pre=NULL;// Record the previous node 
        result.clear();

        searchBST(root);
        return result;

    }
};