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 / 2return [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). 渣方法
/** * 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: map<int, int> ans; void dfs(TreeNode* node){ if(node == NULL) return; ans[node->val]++; dfs(node->left); dfs(node->right); } vector<int> findMode(TreeNode* root) { dfs(root); vector<int> v; int max = -1; for(auto itr = ans.begin(); itr != ans.end(); ++itr){ if(itr->second > max){ max = itr->second; v.clear(); v.push_back(itr->first); } else if(itr->second == max){ v.push_back(itr->first); } } return v; }};新闻热点
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