目录
前言
LeetCode235. 二叉搜索树的最近公共祖先
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
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
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;
}
};