代码随想录算法训练营day20、21 | 701.二叉搜索树中的插入操作、450.删除二叉搜索树中的节点、669. 修剪二叉搜索树、108.有序数组转换为二叉搜索树、538.把二叉搜索树转换为累加树

701.二叉搜索树中的插入操作

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class Solution {
public:
    void preOrder(TreeNode *root, TreeNode *parent, int val) {
        if(root == nullptr) {
            TreeNode *newNode = new TreeNode(val);
            if(val < parent->val) parent->left = newNode;
            else parent->right = newNode;
            return;
        }
        if(root->val < val) preOrder(root->right, root, val);
        if(root->val > val) preOrder(root->left, root, val);
    }
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if(root == nullptr) return new TreeNode(val);
        preOrder(root, nullptr, val);
        return root;
    }
};

450.删除二叉搜索树中的节点
有几个特殊用例通过不了，先跳过...

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class Solution {
public:
    TreeNode *keypre;
    TreeNode *searchKey(TreeNode* root, int key) {
        if(root == nullptr) return nullptr;
        else if(root->val == key) return root;
        else if(root->val > key) {
            keypre = root;
            return searchKey(root->left, key);
        }
        else {
            keypre = root;
            return searchKey(root->right, key);
        }
    }

    void findKeyPreoreNext(TreeNode* root) {
        TreeNode *pre = root;
        TreeNode *cur = root;
        if(root->left != nullptr) {
            cur = root->left;
            while(cur->right != nullptr) {
                pre = cur;
                cur = cur->right;
            }
            swap(root->val, cur->val);
            if(cur == pre->right) pre->right = nullptr;
            if(cur == pre->left) pre->left = cur->left;
        }
        else if(root->right != nullptr) {
            cur = root->right;
            while(cur->left != nullptr) {
                pre = cur;
                cur = cur->left;
            }
            swap(root->val, cur->val);
            if(cur == pre->right) pre->right = cur->right;
            if(cur == pre->left) pre->left = nullptr;
        }
    }
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(root == nullptr) return root;
        if(root->val == key && root->left == nullptr && root->right == nullptr)
            return nullptr;
        if(searchKey(root, key) == nullptr) return root;
        else if(searchKey(root, key)->left == nullptr && searchKey(root, key)->right == nullptr) {
            if(searchKey(root, key) == keypre->left) keypre->left = nullptr;
            if(searchKey(root, key) == keypre->right) keypre->right = nullptr;
            return root;
        }
        else findKeyPreoreNext(searchKey(root, key));
        return root;
    }
};

69.修剪二叉搜索树
难，跳过

108.将有序数组转换为二叉搜索树
难，跳过

538.把二叉搜索树转换为累加树

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class Solution {
public:
    TreeNode *pre = nullptr;
    void inOrder(TreeNode *root) {
        if(root == nullptr) return;
        inOrder(root->right);
        if(pre != nullptr) root->val += pre->val; //第一个节点值最大，无需累加
        pre = root;
        inOrder(root->left);
    }
    TreeNode* convertBST(TreeNode* root) {
        inOrder(root);
        return root;
    }
};

通过右 中 左遍历，从元素最大的节点开始处理，一直处理到元素最小的节点，逐步累加节点值即可

2025/03/02