代码随想录算法训练营第17天|530.二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

530.二叉搜索树的最小绝对差

题目链接

提交代码

class Solution {
public:TreeNode* pre = nullptr;int result = INT_MAX;void traversal(TreeNode* node){if(node == nullptr) return;traversal(node -> left);if(pre != nullptr){if(abs(node -> val - pre -> val) < result)result = abs(node -> val - pre -> val);}          pre = node;traversal(node -> right);}int getMinimumDifference(TreeNode* root) {TreeNode* cur = root;traversal(root);return result;}
};

提交代码（二分法）

class Solution {
public:vector<int> inorder;void traversal(TreeNode* node){if(node == nullptr) return;traversal(node -> left);inorder.push_back(node -> val);traversal(node -> right);}int getMinimumDifference(TreeNode* root) {traversal(root);int result = INT_MAX;for(int i = 1; i < inorder.size(); i++){if(abs(inorder[i] - inorder[i - 1]) < result)result = abs(inorder[i] - inorder[i - 1]);}return result;}
};

501.二叉搜索树中的众数

题目链接

提交代码（方法）

class Solution {
public:vector<int> result;int max_count = 0;int count = 0;TreeNode* pre = nullptr;void traversal(TreeNode* node){if(node == nullptr) return;traversal(node -> left);if(pre){if(node -> val == pre -> val)count++;elsecount = 1;if(count > max_count){result.clear();result.push_back(node -> val);max_count = count;}else if(count == max_count)result.push_back(node -> val);}else{count++;result.push_back(node -> val);max_count = count;}pre = node;traversal(node -> right);}vector<int> findMode(TreeNode* root) {traversal(root);return result;}
};

236.二叉树的最近公共祖先

题目链接

提交代码（方法）

class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == p || root == q || root == nullptr) return root;TreeNode* left = lowestCommonAncestor(root -> left, p, q);TreeNode* right = lowestCommonAncestor(root -> right, p ,q);if(left != nullptr && right != nullptr) return root;else if(left != nullptr && right == nullptr) return left;else if(left == nullptr && right != nullptr) return right;return nullptr;}
};

总结

                     日期： 2023 年 3 月 15 日
              学习时长： 1 h 0 m
                     难度：$★$ $★$
累计完成题目数量： 57
距离目标还有数量： 243
                      小结：