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;
    }
};