530.二叉搜索树的最小绝对差

//耗时巨长写法

int min = INT_MAX;int getMinimumDifference(TreeNode* root) {queue<TreeNode*> qe;if(root == nullptr){return 0;}qe.push(root);vector<vector<int>> result;while(!qe.empty()){int sz = qe.size();vector<int> tmp;for(int i = 0; i < sz; i++){TreeNode* node = qe.front();qe.pop();tmp.push_back(node->val);if(node->left) qe.push(node->left);if(node->right) qe.push(node->right);}result.push_back(tmp);}for(int j = 0; j < result.size()-1; j++){vector<int> pre = result[j];for(int k = j+1; k < result.size(); k++){vector<int> now = result[k];for(int m = 0; m < pre.size(); m++){int tmp_pre = pre[m];for(int n = 0; n < now.size(); n++){int tmp_now = now[n];if(abs(tmp_pre-tmp_now) < min){min = abs(tmp_pre-tmp_now);}}}}}return min;}

//理解二叉搜索树的含义，以及双指针法记录前序指针的写法

    int result = INT_MAX;TreeNode* pre = nullptr;void traverse(TreeNode* node){if(node == nullptr){return;}traverse(node->left);if(pre){result = min(result, abs(node->val - pre->val));}pre = node;traverse(node->right);}int getMinimumDifference(TreeNode* root) {traverse(root);return result;}

501.二叉搜索树中的众数，需二刷，理解双指针及如何求众数

    vector<int> result;TreeNode* prev;int max_count = 0;int count = 0;void traverse(TreeNode* node){if(node == nullptr) return;traverse(node->left);if(prev == nullptr){count = 1;}else if(prev->val == node->val){count += 1;}else{count = 1;}prev = node;if(count > max_count){max_count = count;result.clear();result.push_back(node->val);}else if (count == max_count){result.push_back(node->val);}traverse(node->right);}vector<int> findMode(TreeNode* root) {traverse(root);return result;}

236.二叉树的最长公共祖先，需二刷

//左右中后序遍历，回溯来实现由底向上遍历，找到公共祖先

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == nullptr){return nullptr;}if(root == p || root == q){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 right;}else if(left != nullptr && right == nullptr){return left;}else{return nullptr;}}