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).

 

这道题求二分搜索树中的众数，这里定义的二分搜索树中左根右结点之间的关系是小于等于的，可以利用一个哈希表来记录数字和其出现次数之前的映射，然后维护一个变量mx来记录当前最多的次数值，这样在遍历完树之后，根据这个mx值就能把对应的元素找出来。

class Solution {
public:
    vector<int> findMode(TreeNode* root) {
        if (!root) return {};
        vector<int> res;
        TreeNode *p = root;
        stack<TreeNode*> s;
        unordered_map<int, int> m;
        int mx = 0;
        while (!s.empty() || p) {
            while (p) {
                s.push(p);
                p = p->left;
            }
            p = s.top(); s.pop();
            mx = max(mx, ++m[p->val]);
            p = p->right;
        }
        for (auto a : m) {
            if (a.second == mx) {
                res.push_back(a.first);
            }
        }
        return res;
    }
};