【谷歌面试题】找出二叉查找树中出现频率最高的元素
找出二叉查找树中出现频率最高的元素。树中结点满足left->val <= root->val <= right->val。如果多个元素出现次数相等,返回最小的元素。
在一个有序数组中,我们查找出现频率最高的元素,很简单,顺序扫描一遍即可统计出。那么我们对二叉查找树也可以用类似方式统计,因为中序遍历序列就是有序序列,所以我们在中序遍历的过程中就可以统计出出现频率最高的元素。
- class TreeNode
- {
- public:
- int val;
- TreeNode *left;
- TreeNode *right;
- TreeNode(int val, TreeNode* left = NULL, TreeNode *right = NULL)
- {
- this->val = val;
- this->left = left;
- this->right = right;
- }
- };
- int GetMostFrequently(TreeNode * root)
- {
- void _GetMostFrequently(TreeNode *root, int & current, int & currentFrequency,
- int & maxFrequency, int & mostFrequently);
- if(root == NULL)
- throw new invalid_argument("Can't be a NULL tree");
- int mostFrequently = INT_MAX;
- int current = INT_MAX;
- int currentFrequency = 0;
- int maxFrequency = 0;
- _GetMostFrequently(root, current, currentFrequency, maxFrequency, mostFrequently);
- return mostFrequently;
- }
- void _GetMostFrequently(TreeNode *root, int & current, int & currentFrequency,
- int & maxFrequency, int & mostFrequently)
- {
- if(root == NULL)
- return;
- _GetMostFrequently(root->left, current, currentFrequency,
- maxFrequency, mostFrequently);
- if(root->val == current)
- ++currentFrequency;
- else
- {
- current = root->val;
- currentFrequency = 1;
- }
- if(currentFrequency > maxFrequency)
- {
- maxFrequency = currentFrequency;
- mostFrequently = current;
- }
- _GetMostFrequently(root->right, current, currentFrequency,
- maxFrequency, mostFrequently);
- }