501. Find Mode in Binary Search Tree

Posted 2021-02-14 simplepaul

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：

对于本题中的二叉搜索树，节点的排列是有顺序的，左节点<=当前节点<=右节点。也就是说，假设这样一种情况：存在大于等于3个的连续相同的节点，且当前节点存在左右子树，那么相同的三个节点一定是：当前节点、左子树中最大的节点和右子树中最小的节点。

这里我们容易想到的是中序遍历（对于整个树，先遍历左节点，再遍历中间节点，最后遍历右节点），使用中序遍历遍历二叉搜索树，可以获得一个从小到大的排好序的升序序列。

所以分析到这里，题目就转化成：给定一个升序序列，寻找重复次数最多的数字 ， 所以

/**
 * 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) {
        tmp_cnt = 0;
        max_cnt = 0;
        cur_value = 0;
        inorder(root);
        return result;
    }
private:
    int tmp_cnt;
    int max_cnt;
    int cur_value;
    vector<int> result;
    void inorder(TreeNode* root){
        if (root == NULL){
            return;
        }
        // 遍历左子树
        inorder(root->left);
        tmp_cnt++;  
        if (cur_value != root-> val){
            cur_value = root-> val;
            tmp_cnt = 1;
        }
        if (tmp_cnt > max_cnt){
            max_cnt = tmp_cnt;
            result.clear();
            result.push_back(root->val);
        }else if (tmp_cnt == max_cnt){
            result.push_back(root->val);
        }
        // 遍历右子树
        inorder(root->right);
    }
};