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：

對於本題中的二叉搜索樹，節點的排列是有順序的，左節點<=當前節點<=右節點。也就是說，假設這樣一種狀況：存在大於等於3個的連續相同的節點，且當前節點存在左右子樹，那麼相同的三個節點必定是：當前節點、左子樹中最大的節點和右子樹中最小的節點。

這裏咱們容易想到的是中序遍歷（對於整個樹，先遍歷左節點，再遍歷中間節點，最後遍歷右節點），使用中序遍歷遍歷二叉搜索樹，能夠得到一個從小到大的排好序的升序序列。

因此分析到這裏，題目就轉化成：給定一個升序序列，尋找重複次數最多的數字 ， 因此

/**
 * 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;
        }
        // 遍歷左子樹
        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);
        }
        // 遍歷右子樹
        inorder(root->right);
    }
};