代码随想录算法训练营第二十天|LeetCode235. 二叉搜索树的最近公共祖先、LeetCode701.二叉搜索树中的插入操作、LeetCode450.删除二叉搜索树中的节点

目录

前言

一、LeetCode235. 二叉搜索树的最近公共祖先

题目链接:

代码:

递归法:

迭代法:

二、LeetCode701.二叉搜索树中的插入操作

题目链接:

代码:

递归法:

迭代法:

三、LeetCode450.删除二叉搜索树中的节点

题目链接:

代码:


前言

LeetCode235. 二叉搜索树的最近公共祖先

文章讲解:代码随想录

视频讲解:LeetCode235. 二叉搜索树的最近公共祖先_哔哩哔哩_bilibili

LeetCode701.二叉搜索树中的插入操作

文章讲解:代码随想录

视频讲解:LeetCode:701.二叉搜索树中的插入操作_哔哩哔哩_bilibili

LeetCode450.删除二叉搜索树中的节点

文章讲解:代码随想录

视频讲解:LeetCode:450.删除二叉搜索树中的节点_哔哩哔哩_bilibili


一、LeetCode235. 二叉搜索树的最近公共祖先

题目链接:

235. 二叉搜索树的最近公共祖先 - 力扣(LeetCode)

代码:

递归法:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == nullptr)
            return nullptr;
        
        if(root->val > p->val && root->val > q->val)
        {
            TreeNode* left = lowestCommonAncestor(root->left, p, q);
            if(left != nullptr)
                return left;
        }

        if(root->val < p->val && root->val < q->val)
        {
            TreeNode* right = lowestCommonAncestor(root->right, p, q);
            if(right != nullptr)
                return right;
        }

        return root;
    }
};

迭代法:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        while(1)
        {
            if(root->val > p->val && root->val > q->val)
                root = root->left;
            else if(root->val < p->val && root->val < q->val)
                root = root->right;
            else
                return root;
        }
    }
};

二、LeetCode701.二叉搜索树中的插入操作

题目链接:

701. 二叉搜索树中的插入操作 - 力扣(LeetCode)

代码:

递归法:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if(root == nullptr)
        {
            return new TreeNode(val);
        }
        TreeNode* pCur = root;
        while(1)
        {
            if(pCur->val > val)
            {
                if(pCur->left == nullptr)
                {
                    TreeNode* pNew = new TreeNode(val);
                    pCur->left = pNew;
                    break;
                }
                else
                {
                    pCur = pCur->left;
                }
            }
            else
            {
                if(pCur->right == nullptr)
                {
                    TreeNode* pNew = new TreeNode(val);
                    pCur->right = pNew;
                    break;
                }
                else
                {
                    pCur = pCur->right;
                }
            }
        }
        return root;
    }
};

迭代法:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if(root == nullptr)
        {
            return new TreeNode(val);
        }
        TreeNode* pCur = root;
        while(1)
        {
            if(pCur->val > val)
            {
                if(pCur->left == nullptr)
                {
                    TreeNode* pNew = new TreeNode(val);
                    pCur->left = pNew;
                    break;
                }
                else
                {
                    pCur = pCur->left;
                }
            }
            else
            {
                if(pCur->right == nullptr)
                {
                    TreeNode* pNew = new TreeNode(val);
                    pCur->right = pNew;
                    break;
                }
                else
                {
                    pCur = pCur->right;
                }
            }
        }
        return root;
    }
};

三、LeetCode450.删除二叉搜索树中的节点

题目链接:

450. 删除二叉搜索树中的节点 - 力扣(LeetCode)

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(root == nullptr) return nullptr;
        if(root->val == key)
        {
            if(root->left == nullptr && root->right == nullptr)
            {
                delete root;
                return nullptr;
            }
            else if(root->left != nullptr && root->right == nullptr)
            {
                TreeNode* ret = root->left;
                delete root;
                return ret;
            }
            else if(root->left == nullptr && root->right != nullptr)
            {
                TreeNode* ret = root->right;
                delete root;
                return ret;
            }
            else // root->left != nullptr && root->right != nullptr
            {
                TreeNode* ret = root->right;
                TreeNode* pCur = root->right;
                while(pCur->left != nullptr)
                {
                    pCur = pCur->left;
                }
                pCur->left = root->left;
                delete root;
                return ret;
            }
        }
        if(root->val > key)
            root->left = deleteNode(root->left, key);
        if(root->val < key)
            root->right = deleteNode(root->right, key);
        
        return root;
    }
};
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