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问题描述：

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.

示例：

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.

问题分析：

可以通过遍历统计各个值出现的次数，最终确定答案。

过程详见代码：

/**
 * 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:
	map<int,int> m;
	int maxi;
public:
    vector<int> findMode(TreeNode* root) {
    	vector<int> res;
        maxi = 0;
        travarse(m,root,maxi);
        for(map<int,int>::iterator it = m.begin(); it != m.end();it++)
        {
        	if(it->second == maxi) res.push_back(it->first);
		}
		return res;
    }
    
    void travarse(map<int,int>& m,TreeNode* root,int& maxi)
    {
    	if(root == NULL) return;
    	if(m.count(root->val) == 0) m.insert(pair<int,int>(root->val,1));
    	else m[root->val]++;
    	if(m[root->val] > maxi) maxi = m[root->val];
    	travarse(m,root->left,maxi);
    	travarse(m,root->right,maxi);
	}
};