1.题目

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],

return [2].

给一棵二叉搜索树，有重复元素。找出其中重复次数最多的元素。如果有多个重复次数一样的元素，则放在一个列表中作为返回结果。

2.分析

这题考察的是BST的中序遍历，中序遍历的元素是有序的，在遍历的过程中统计元素出现的次数即可。

3.代码

中序遍历–迭代

/**
 * 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) {
        vector<int> mode;
        if (root == NULL)
            return mode;
        int max_c = -1;//记录历史最多出现次数
        int last_v = INT_MIN;
        int count = 1;//当前值出现的次数
        stack<TreeNode*> nodes;
        while (!nodes.empty() || root) {
            if (root) {
                nodes.push(root);
                root = root->left;//先遍历到左子树最下边
            }
            else {
                root = nodes.top();
                nodes.pop();
                if (last_v != INT_MIN) {
                    if (root->val == last_v)
                        ++count;
                    else
                        count = 1;
                }
                if (count > max_c)
                {
                    max_c = count;
                    mode.erase(mode.begin(), mode.end());
                    mode.push_back(root->val);
                }
                else if (count == max_c)
                    mode.push_back(root->val);
                last_v = root->val;
                root = root->right;//转到右子树
            }
        }
        return mode;
    }
};

中序遍历–递归

class Solution {
public:
    void inOrder(TreeNode* root, vector<int>&mode,int& val,int& count,int& maxc) {
        if (root == NULL)
            return;
        inOrder(root->left,mode,val,count,maxc);
        if (val != INT_MIN) {
            if (val == root->val)
                ++count;
            else
                count = 1;
        }
        if (count > maxc) {
            maxc = count;
            mode.erase(mode.begin(), mode.end());//出现次数更多的元素，清空容器
            mode.push_back(root->val);
        }
        else if (count == maxc)
            mode.push_back(root->val);
        val = root->val;
        inOrder(root->right, mode, val, count, maxc);
    }
    vector<int> findMode(TreeNode* root) {
        vector<int> mode;
        if (root == NULL)
            return mode;
        int maxc = -1;
        int count = 1;
        int val = INT_MIN;
        inOrder(root, mode, val, count, maxc);
        return mode;
    }
};