文章作者:Tyan
博客:noahsnail.com | 优快云 | 简书
1. Description

2. Solution
/**
* 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> modes;
if(!root) {
return modes;
}
int count = 0;
int max = 0;
int prev = 0;
inorder(root, modes, count, max, prev);
return modes;
}
private:
void inorder(TreeNode* root, vector<int>& modes, int& count, int& max, int& prev) {
if(!root) {
return;
}
inorder(root->left, modes, count, max, prev);
if(root->val == prev) {
count++;
}
else {
count = 1;
}
if(count > max){
modes.clear();
max = count;
modes.push_back(root->val);
}
else if(count == max) {
modes.push_back(root->val);
}
prev = root->val;
inorder(root->right, modes, count, max, prev);
}
};
本文介绍了一种使用中序遍历算法在二叉搜索树中查找出现频率最高的元素的方法。通过一次遍历,算法可以高效地找到众数,并在代码中详细解释了其实现过程。
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