235. 二叉搜索树的最近公共祖先

和昨天的一题很像，两种做法
1、dfs

class Solution {
public:
    // dfs版
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == nullptr || 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 nullptr;
        if (left == nullptr) return right;
        if (right == nullptr) return left;
        return root;
    }
};

2、利用二叉搜索树的特性

class Solution {
public:
    // 不需要dfs,普通遍历即可
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == nullptr || root == p || root == q) return root;
        if (p->val > q->val) swap(p, q);
        while (root != nullptr)
        {
            if (root->val < q->val && root->val > p->val)
                return root;
            if (root->val == p->val || root->val == q->val)
                return root;
            if (root->val > p->val)
                root = root->left;
            else
                root = root->right;
        }
        return nullptr;
    }
};

701. 二叉搜索树中的插入操作

找到的点，一定有位置，直接插入即可。

class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == nullptr) return new TreeNode(val);
        TreeNode* work = root;
        TreeNode* pre = root;
        while (work != nullptr)
        {
            pre = work;
            if (work->val < val) work = work->right;
            else if (work->val > val) work = work->left;
        }
        TreeNode* res = new TreeNode(val);
        if (val < pre->val) 
            pre->left = res;
        else pre->right = res;
        return root;
    }
};

450. 删除二叉搜索树中的节点

五种情况：没搜到；叶子节点；左空；右空；都不为空

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (root == nullptr) return nullptr;
        if (key > root->val) root->right = deleteNode(root->right, key); // 去右子树
        else if (key < root->val) root->left = deleteNode(root->left, key); // 去左子树
        else
        {
            if (!root->left) return root->right; // 待删除的节点没有左孩子，直接把右孩子返回给上一级
            if (!root->right) return root->left; // 没有右孩子
            // 找到右孩子的左下
            TreeNode* node = root->right;
            while (node->left)
                node = node->left;
            node->left = root->left;
            root = root->right;
        }
        return root;
    }
};