算法训练营day24
题目1：235. 二叉搜索树的最近公共祖先 - 力扣（LeetCode）
可以用二叉树最近公共祖先来做
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(leftnode != NULL && rightnode != NULL) return root;else if(leftnode != NULL && rightnode == NULL) return leftnode;else if(leftnode == NULL && rightnode != NULL) return rightnode;else return NULL;}
};class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(root == NULL) return NULL;// p q都在 node左边if(root->val > p->val && root->val > q->val) {TreeNode* leftnode = lowestCommonAncestor(root->left, p, q);if(leftnode != NULL) return leftnode;}// p q都在 node右边if(root->val < p->val && root->val < q->val) {TreeNode* rightnode = lowestCommonAncestor(root->right, p, q);if(rightnode != NULL) return rightnode;}// 就剩在中间了return root;}
};题目2：701. 二叉搜索树中的插入操作 - 力扣（LeetCode）
// 根据搜索二叉树的性质，向左向右遍历遇到NULL 就插入节点即可
class Solution {
public:void insert(TreeNode*& root, int val) {if(root == NULL) {TreeNode* node = new TreeNode(val);root = node;return;}if(root->val > val) {insert(root->left, val);}if(root->val < val) {insert(root->right, val);}}TreeNode* insertIntoBST(TreeNode* root, int val) {insert(root, val);return root; }
};题目3：450. 删除二叉搜索树中的节点 - 力扣（LeetCode）
class Solution {
public:TreeNode* deleteNode(TreeNode* root, int key) {if(root == NULL) return root;if(root->val == key) {if(root->left == NULL && root->right == NULL) {delete root;return NULL;}else if(root->left == NULL && root->right != NULL) {TreeNode* tmp = root->right;delete root;return tmp;}else if(root->left != NULL && root->right == NULL) {TreeNode* tmp = root->left;delete root;return tmp;}else {TreeNode* cur = root->right;while(cur->left != NULL) cur = cur->left;cur->left = root->left;TreeNode* tmp = root->right;delete root;return tmp;}}if(root->val > key) root->left = deleteNode(root->left, key);if(root->val < key) root->right = deleteNode(root->right, key);return root;}
};