算法训练营day23补签

 题目1：530. 二叉搜索树的最小绝对差 - 力扣（LeetCode）

class Solution {
public:int reslut = INT_MAX;TreeNode* pre = NULL;void trackingback(TreeNode* node) {if(node == NULL) return;trackingback(node->left);if(pre != NULL) {reslut = min(reslut, node->val - pre->val);}pre = node;trackingback(node->right);}  int getMinimumDifference(TreeNode* root) {trackingback(root);return reslut;}
};

 题目2：501. 二叉搜索树中的众数 - 力扣（LeetCode）

// 中序遍历用map记录每个数出现的个数，然后找最大的
class Solution {
public:unordered_map<int, int> map;void trackingback(TreeNode* node) {if(node == NULL) return;trackingback(node->left);if(map.find(node->val) != map.end()) {map[node->val]++;}else {map.insert(pair<int, int>(node->val, 1));}trackingback(node->right);}vector<int> findMode(TreeNode* root) {vector<int> reslut;trackingback(root);int num = 1;for(auto iter =map.begin();iter != map.end();iter++) {if(iter->second > num) {num = iter->second;}}for(auto iter =map.begin();iter != map.end();iter++) {if(iter->second == num) {reslut.push_back(iter->first);}}return reslut;}
};

在递归里操作，主要就是更新maxcount 然后要把reslut也更新 

class Solution {
public:vector<int> reslut;int count = 0;int maxcount = 0;TreeNode* pre = NULL;void trackingback(TreeNode* node) {if(node == NULL) return;trackingback(node->left);if(pre == NULL) {count = 1;}else if(pre->val == node->val) {count++ ;}else if(pre->val != node->val) {count = 1;}pre = node;if(count == maxcount) {reslut.push_back(node->val);}if(count > maxcount) {reslut.clear();maxcount = count;reslut.push_back(node->val);}trackingback(node->right);}vector<int> findMode(TreeNode* root) {trackingback(root);return reslut;}
};

 题目3：236. 二叉树的最近公共祖先 - 力扣（LeetCode）

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