原题：
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).
代码如下：
/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ /** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* findMode(struct TreeNode* root, int* returnSize) { if(root==NULL) return root; int* max_times; int* max_num; int* flag_num; int* flag_times; max_times=(int*)malloc(sizeof(int)); max_num=(int*)malloc(sizeof(int)*20000); flag_num=(int*)malloc(sizeof(int)); flag_times=(int*)malloc(sizeof(int)); *max_times=0; *flag_times=0; void findmaxtimes(struct TreeNode* ,int* ,int* ,int* ,int*,int*); findmaxtimes(root,max_times,max_num,flag_num,flag_times,returnSize); if(*flag_times>*max_times) { *returnSize=1; return flag_num; } if(*flag_times==*max_times) { *(max_num+*returnSize)=*flag_num; *returnSize+=1; } return max_num; } void findmaxtimes(struct TreeNode* root,int* max_times,int* max_num,int* flag_num,int* flag_times,int* returnSize) { if(root==NULL) return; findmaxtimes(root->left,max_times,max_num,flag_num,flag_times,returnSize); if(*flag_times==0) *flag_num=root->val; if(root->val==*flag_num) *flag_times+=1; else { if(*flag_times==*max_times) { *(max_num+*returnSize)=*flag_num; *returnSize+=1; } if(*flag_times>*max_times) { *max_times=*flag_times; *max_num=*flag_num; *returnSize=1; } *flag_num=root->val; *flag_times=1; } findmaxtimes(root->right,max_times,max_num,flag_num,flag_times,returnSize); } 思路其实不难，就是前序遍历，生成出来一个有序数列，然后统计频率就好。但是要完成他的要求，就要求在遍历过程中完成统计和比较，其实耐心一点不难写。（我第一次写的时候就觉得贼烦。。。）
