Sunday, November 29, 2015

Google Interview - 找出二叉查找树中出现频率最高的元素 - 我的博客 - ITeye技术网站


Google Interview - 找出二叉查找树中出现频率最高的元素 - 我的博客 - ITeye技术网站
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。
  1. int GetMostFrequently(TreeNode *root) {    
  2.     if(root == NULL)    
  3.         throw new invalid_argument("Can't be a NULL tree");    
  4.     int mostFrequently = INT_MAX;    
  5.     int current = INT_MAX;    
  6.     int currentFrequency = 0;    
  7.     int maxFrequency = 0;    
  8.     _GetMostFrequently(root, current, currentFrequency, maxFrequency, mostFrequently);    
  9.     return mostFrequently;    
  10. }    
  11.     
  12. void _GetMostFrequently(TreeNode *root, int &current, int & currentFrequency, int &maxFrequency, int &mostFrequently)  {    
  13.     if(root == NULL)  return;    
  14.     _GetMostFrequently(root->left, current, currentFrequency,     
  15.                 maxFrequency, mostFrequently);    
  16.     if(root->val == current)  {  
  17.         ++currentFrequency;    
  18.     } else  {    
  19.         current = root->val;    
  20.         currentFrequency = 1;    
  21.     }    
  22.     if(currentFrequency > maxFrequency)  {    
  23.         maxFrequency = currentFrequency;    
  24.         mostFrequently = current;    
  25.     }    
  26.     _GetMostFrequently(root->right, current, currentFrequency, maxFrequency, mostFrequently);    
  27. }   

  1. private static class Stat {  
  2.     int freq, val;  
  3.     boolean inited;  
  4. }  
  5.   
  6. public int mostFrequent(TreeNode root) throws Exception {  
  7.     if(root == nullthrow new Exception("null root!");  
  8.     Stat result = new Stat(), current = new Stat();  
  9.     mostFrequent(root, result, current);  
  10.     return result.val;  
  11. }  
  12.   
  13. private void mostFrequent(TreeNode node, Stat result, Stat current) {  
  14.     if(node == nullreturn;  
  15.     mostFrequent(node.left, result, current);  
  16.     if(!current.inited || node.val != current.val) {  
  17.         current.inited = true;  
  18.         current.freq = 1;  
  19.         current.val = node.val;  
  20.     } else {  
  21.         current.freq++;  
  22.     }  
  23.     if(current.freq > result.freq) {  
  24.         result.freq = current.freq;  
  25.         result.val = current.val;  
  26.     } else if(current.freq == result.freq && current.val < result.val) {  
  27.         result.val = current.val;  
  28.     }  
  29.     mostFrequent(node.right, result, current);  
  30. }  
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