530. 二叉搜索树的最小绝对差
1. LeetCode链接
力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台
2. 题目描述
3. 解法
中序遍历，记录前一个指针，并记录前一个指针和当前指针的绝对差值。递归。
class Solution {
public:TreeNode* pre = NULL;int min = INT_MAX;void order(TreeNode* root) {if (root == NULL) return;order(root->left);if (pre != NULL && root->val - pre->val < min) {min = root->val - pre->val;}pre = root;order(root->right);}int getMinimumDifference(TreeNode* root) {order(root);return min;}
};统一迭代
class Solution {
public:int getMinimumDifference(TreeNode* root) {TreeNode* pre = NULL;int result = INT_MAX;stack<TreeNode*> st;st.push(root);while (!st.empty()) {TreeNode* cur = st.top();if (cur != NULL) {st.pop();if (cur->right != NULL) st.push(cur->right);st.push(cur);st.push(NULL);if (cur->left != NULL) st.push(cur->left);} else {st.pop();if (pre != NULL && (st.top()->val - pre->val) < result) result = st.top()->val - pre->val;pre = st.top();st.pop();}}return result;}
};501. 二叉搜索树中的众数
1. LeetCode链接
力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台
2. 题目描述
3. 解法
中序遍历，记录指针、最大出现频率、当前数字累计个数、最终result。
class Solution {
public:TreeNode* pre = NULL;int max = 1;int count = 1;vector<int> result;void order(TreeNode* root) {if (root == NULL) return;order(root->left);if (pre != NULL && root->val == pre->val) count++;if (pre != NULL && root->val != pre->val) count = 1;pre = root;if (count > max) {result.erase(result.begin(), result.end());result.push_back(pre->val);max = count;} else if (count == max) result.push_back(pre->val);order(root->right);}vector<int> findMode(TreeNode* root) {order(root);return result;}
};236. 二叉树的最近公共祖先
1. LeetCode链接
力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台
2. 题目描述
3. 解法
自己想到的笨办法，自顶向下找，每次都要遍历一遍当前节点之下的节点。很耗时。
class Solution {
public:TreeNode* result;bool exist(TreeNode* root, TreeNode* p) {if (root == NULL) return false;if (root == p) return true;bool left = exist(root->left, p);bool right = exist(root->right, p);return left || right;}void order(TreeNode* root, TreeNode* p, TreeNode* q) {if (root == NULL) return;if (exist(root, p) && exist(root, q)) result = root;if (root == p || root == q) return;order(root->left, p, q);order(root->right, p, q);}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {order(root, p, q);return result;}
};自底向上递归。就是找到p、q节点。如果，恰好左右节点分别在某节点的左右子树上，直接返回这个节点，即为公共节点。
从下往上遍历，就用后序遍历，先判断完左右子树，然后根据结果判断当前节点。
class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if (root == p || root == q || 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 && right == NULL) return left;else if (left == NULL && right != NULL) return right;else return NULL;}
};
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