标签：ESS vector ace 思路 using https 搜索 equal col
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:
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).
思路：
对于本题中的二叉搜索树，节点的排列是有顺序的，左节点<=当前节点<=右节点。也就是说，假设这样一种情况：存在大于等于3个的连续相同的节点，且当前节点存在左右子树，那么相同的三个节点一定是：当前节点、左子树中最大的节点和右子树中最小的节点。
这里我们容易想到的是中序遍历（对于整个树，先遍历左节点，再遍历中间节点，最后遍历右节点），使用中序遍历遍历二叉搜索树，可以获得一个从小到大的排好序的升序序列。
所以分析到这里，题目就转化成：给定一个升序序列，寻找重复次数最多的数字 ， 所以
/** * 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) { tmp_cnt = 0; max_cnt = 0; cur_value = 0; inorder(root); return result; } private: int tmp_cnt; int max_cnt; int cur_value; vector<int> result; void inorder(TreeNode* root){ if (root == NULL){ return; } // 遍历左子树 inorder(root->left); tmp_cnt++; if (cur_value != root-> val){ cur_value = root-> val; tmp_cnt = 1; } if (tmp_cnt > max_cnt){ max_cnt = tmp_cnt; result.clear(); result.push_back(root->val); }else if (tmp_cnt == max_cnt){ result.push_back(root->val); } // 遍历右子树 inorder(root->right); } };
501. Find Mode in Binary Search Tree
标签：ESS vector ace 思路 using https 搜索 equal col
原文地址：https://www.cnblogs.com/simplepaul/p/10569051.html
