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代码随想录算法训练营第37期 第二十一天 | LeetCode530.二叉搜索树的最小绝对差、501.二叉搜索树中的众数、236. 二叉树的最近公共祖先
一、530.二叉搜索树的最小绝对差
解题代码C++：
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
private:int result = INT_MAX;TreeNode* pre = NULL;void traversal(TreeNode* cur){if(cur == NULL) return;traversal(cur->left);if(pre != NULL)result = min(result, cur->val - pre->val);pre = cur;traversal(cur->right);}public:int getMinimumDifference(TreeNode* root) {traversal(root);return result;}
};
题目链接/文章讲解/视频讲解：
https://programmercarl.com/0530.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E6%9C%80%E5%B0%8F%E7%BB%9D%E5%AF%B9%E5%B7%AE.html
二、501.二叉搜索树中的众数
解题代码C++：
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
private:int maxCount = 0;int count = 0;TreeNode* pre = NULL;vector<int> result;void searchBST(TreeNode* cur){if(cur == NULL) return;searchBST(cur->left);if(pre == NULL)count = 1;else if(pre->val == cur->val)count ++;elsecount = 1;pre = cur;if(count == maxCount)result.push_back(cur->val);if(count > maxCount){maxCount = count;result.clear();result.push_back(cur->val);}searchBST(cur->right);return;}public:vector<int> findMode(TreeNode* root) {count = 0;maxCount = 0;pre = NULL;result.clear();searchBST(root);return result;}
};
题目链接/文章讲解/视频讲解：
https://programmercarl.com/0501.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E4%BC%97%E6%95%B0.html
三、236. 二叉树的最近公共祖先
解题代码C++：
/*** 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:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == q || root == p || root == NULL) return root;TreeNode* left = lowestCommonAncestor(root->left, p, q);TreeNode* right = lowestCommonAncestor(root->right, p, q);if(left != NULL && right != NULL) return root;if(left == NULL) return right;return left;}
};
题目链接/文章讲解/视频讲解：
https://programmercarl.com/0236.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88.html