代码随想录算法训练营DAY20第六章 二叉树part07

目录

[235. 二叉搜索树的最近公共祖先](#235. 二叉搜索树的最近公共祖先)

[701. 二叉搜索树中的插入操作](#701. 二叉搜索树中的插入操作)

[450. 删除二叉搜索树中的节点](#450. 删除二叉搜索树中的节点)

---

235. 二叉搜索树的最近公共祖先

/**
 * 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 (p->val < root->val && q->val < root->val)
            return lowestCommonAncestor(root->left, p, q);
        if (p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right, p, q);
        return root;
    }
};

701. 二叉搜索树中的插入操作

/**
 * 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)return new TreeNode(val);
        if(val<root->val)root->left=insertIntoBST(root->left,val);
        else root->right=insertIntoBST(root->right,val);
        return root;
    }
};

450. 删除二叉搜索树中的节点

/**
 * 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)return nullptr;
        if(key>root->val)root->right=deleteNode(root->right,key);
        else if(key<root->val) root->left=deleteNode(root->left,key);
        else{
            if(!root->left)return root->right;
            else if(!root->right)return root->left;
            TreeNode* node=root->right;
            while(node->left)node=node->left;
            node->left=root->left;
            root=root->right;
        }
        return root;
    }
};