Find Mode in Binary Search Tree【找到BST树中出现最多的元素】

PROBLEM:

---

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).

SOLVE:

/**
 * 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* r) {
      int mfq/*max freq*/, pre/*previous val*/, cnt/*duplicates count*/;
      vector<int> modes;
      getMaxFreq(r, mfq=0, pre, cnt=0); // in-order traversal to get max frequency
      getMode(r, mfq, pre, cnt=0, modes);      // in-order traversal to get all modes
      return modes;
    }
    void getMaxFreq(TreeNode* r, int& mfq, int& pre, int& cnt) {
      if (!r) return;
      getMaxFreq(r->left, mfq, pre, cnt);
      getMaxFreq(r->right, mfq=max(mfq,cnt), pre=r->val, ++(cnt*=(r->val==pre)));
    }
    void getMode(TreeNode* r, const int mfq, int& pre, int& cnt, vector<int>& modes) {
      if (!r) return;
      getMode(r->left, mfq, pre, cnt, modes);
      if (mfq == ++(cnt*=(r->val==pre))) modes.push_back(r->val);
      getMode(r->right, mfq, pre=r->val, cnt, modes);
    }
};

思路：如果是在一个排好序的数组中查找其实很容易，但是题目要求不能使用extra内存，所以就要在树中按从小到大的顺序遍历：方法就是指针先指到最左的一个元素然后遍历右子树（如果存在的话，否则退到上一层）。函数需要储存上一个值，相同值的个数（其实之后把他当成数组遍历就好了）。题解中第一个函数是题目给定的，getMaxFreq（）函数找出出现最多的次数mfq，getMode（）找出出现次数为mfq的元素。

解法中 getMaxFreq(r->left, mfq, pre, cnt) 的作用都是移动到当前子树中的最左子节点， getMaxFreq(r->right, ***) 开始按顺序遍历。

注意变量 mfq，pre，cnt的初始化。