LeetCode

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).

问在一棵BST中，出现最多的数是哪个（哪些）

对BST进行中根遍历的话，结果就刚好是一个排序数组。我们记录他的上一个数字prenum，最大的次数cntmax，和当前已经出现过的次数cntnow。如果cntnow > cntmax，那么久清空数组并更新，如果相等，直接丢进数组就好了。时间复杂度O(n)，空间复杂度我觉得最坏情况下有O(n)，不懂那些说O(1)的是怎么想的。

/** * 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) {        solve(root);        return ans;    }private:    vector<int> ans;    int cntnow = 0, cntmax = 0, prenum = 0;    void solve(TreeNode* root) {        if (!root) return;        solve(root->left);        cntnow++;        if (prenum != root->val) {            prenum = root->val;            cntnow = 1;        }        if (cntnow > cntmax) {            cntmax = cntnow;            ans.clear();            ans.push_back(root->val);        }        else if (cntnow == cntmax) {            ans.push_back(root->val);        }        solve(root->right);    }};