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).

 

求一棵二叉搜索树结点相等值的最大个数

利用BST中序遍历后为排序数组的特点，相等的数是连续的

 

C++（15ms）：

/**
 * 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 {
private:
    TreeNode* prev = NULL ;
    int count = 1;
    int max = 0 ;
public:
    vector<int> findMode(TreeNode* root) {
        vector<int> res ;
        if (root == NULL)
            return res; 
        traverse(root , res) ;
        return res ;
    }
    void traverse(TreeNode* root , vector<int>& res){
        if (root == NULL)
            return ;
        traverse(root->left , res) ;
        if (prev != NULL){
            if (root->val == prev->val)
                count++ ;
            else
                count = 1; 
        }
        if (count > max){
            res.clear() ;
            max = count ;
            res.push_back(root->val) ;
        }else if (count == max){
            res.push_back(root->val) ;
        }
        prev = root ;
        traverse(root->right , res) ;
        
    }
};