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masx200masx200
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https://leetcode.cn/problems/insert-into-a-binary-search-tree/
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‎README.md

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##### 包含的内容如下
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https://leetcode.cn/problems/insert-into-a-binary-search-tree/
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https://leetcode.cn/problems/adding-two-negabinary-numbers/
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https://leetcode.cn/problems/lMSNwu/

‎add-two-integers/export.go

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package add_two_integers
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func Sum(num1 int, num2 int) int {
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return sum(num1, num2)
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}
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package add_two_integers
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func Sum(num1 int, num2 int) int {
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return sum(num1, num2)
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}

‎add-two-integers/index.go

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package add_two_integers
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func sum(num1 int, num2 int) int {
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return num1 + num2
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}
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package add_two_integers
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func sum(num1 int, num2 int) int {
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return num1 + num2
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}

‎insert-into-a-binary-search-tree/export.go

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package insert_into_a_binary_search_tree
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func InsertIntoBST(root *TreeNode, val int) *TreeNode {
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return insertIntoBST(root, val)
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}

‎insert-into-a-binary-search-tree/index.go

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package insert_into_a_binary_search_tree
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import serialize_and_deserialize_binary_tree "github.com/masx200/leetcode-test/serialize-and-deserialize-binary-tree"
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type TreeNode = serialize_and_deserialize_binary_tree.TreeNode
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func insertIntoBST(root *TreeNode, val int) *TreeNode {
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if root == nil {
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root = &TreeNode{Val: val}
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return root
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}
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if root.Val > val {
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root.Left = insertIntoBST(root.Left, val)
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} else {
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root.Right = insertIntoBST(root.Right, val)
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}
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return root
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}

‎insert-into-a-binary-search-tree/index.kt

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package masx200.leetcode_test.insert_into_a_binary_search_tree
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class TreeNode(var `val`: Int = 0, var left: TreeNode? = null, var right: TreeNode? = null) {}
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class Solution {
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fun insertIntoBST(root: TreeNode?, `val`: Int): TreeNode? {
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if (root == null) return TreeNode(`val`)
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if (root.`val` > `val`) root.left = insertIntoBST(root.left, `val`)
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if (root.`val` < `val`) root.right = insertIntoBST(root.right, `val`)
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return root
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}
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}

‎insert-into-a-binary-search-tree/index.ts

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import { TreeNode } from "../binary-tree-inorder-traversal/TreeNode.ts";
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export default function insertIntoBST(
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root: TreeNode | null,
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val: number,
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): TreeNode | null {
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if (root === null) return { val, left: null, right: null };
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if (root.val > val) root.left = insertIntoBST(root.left, val);
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else if (root.val < val) root.right = insertIntoBST(root.right, val);
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return root;
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}

‎insert-into-a-binary-search-tree/readme.md

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701. 二叉搜索树中的插入操作
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https://leetcode.cn/problems/insert-into-a-binary-search-tree/
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简单递归版本
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```java []
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class Solution {
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public TreeNode insertIntoBST(TreeNode root, int val) {
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if (root == null) // 如果当前节点为空，也就意味着val找到了合适的位置，此时创建节点直接返回。
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return new TreeNode(val);
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if (root.val < val){
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root.right = insertIntoBST(root.right, val); // 递归创建右子树
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}else if (root.val > val){
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root.left = insertIntoBST(root.left, val); // 递归创建左子树
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}
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return root;
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}
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}
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```
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```kotlin []
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/**
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* Example:
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* var ti = TreeNode(5)
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* var v = ti.`val`
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* Definition for a binary tree node.
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* class TreeNode(var `val`: Int) {
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* var left: TreeNode? = null
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* var right: TreeNode? = null
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* }
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*/
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class Solution {
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fun insertIntoBST(root: TreeNode?, `val`: Int): TreeNode? {
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if(root==null)return TreeNode(`val`)
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if(root.`val`>`val`)root.left=insertIntoBST(root.left,`val`)
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if(root.`val`<`val`)root.right=insertIntoBST(root.right,`val`)
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return root
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}
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}
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```
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```typescript []
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
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if(root === null) return { val,left:null,right:null }
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if(root.val > val) root.left = insertIntoBST(root.left,val)
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else if(root.val < val) root.right = insertIntoBST(root.right,val);
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return root;
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};
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```
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```go []
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func insertIntoBST(root *TreeNode, val int) *TreeNode {
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if root == nil {
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root = &TreeNode{Val: val}
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return root
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}
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if root.Val > val {
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root.Left = insertIntoBST(root.Left, val)
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} else {
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root.Right = insertIntoBST(root.Right, val)
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}
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return root
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}
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```
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```cpp []
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class Solution {
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public:
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TreeNode* insertIntoBST(TreeNode* root, int val) {
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if (root == NULL) {
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TreeNode* node = new TreeNode(val);
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return node;
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}
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if (root->val > val) root->left = insertIntoBST(root->left, val);
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if (root->val < val) root->right = insertIntoBST(root->right, val);
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return root;
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}
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};
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```
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```php []
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class Solution
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{
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// 递归函数，返回构造后 BST 的根节点
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public function insertIntoBST($root, $val)
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{
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if ($root === null) return new TreeNode($val);
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if ($val < $root->val) {
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$root->left = $this->insertIntoBST($root->left, $val);
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} else {
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$root->right = $this->insertIntoBST($root->right, $val);
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}
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return $root;
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}
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}
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```
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```csharp []
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* public int val;
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* public TreeNode left;
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* public TreeNode right;
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* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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public class Solution {
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public TreeNode InsertIntoBST(TreeNode root, int val) {
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if (root == null) // 如果当前节点为空，也就意味着val找到了合适的位置，此时创建节点直接返回。
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return new TreeNode(val);
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if (root.val < val){
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root.right = InsertIntoBST(root.right, val); // 递归创建右子树
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}else if (root.val > val){
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root.left = InsertIntoBST(root.left, val); // 递归创建左子树
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}
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return root;
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}
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}
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```
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```python []
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class Solution:
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def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
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if root is None:
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return TreeNode(val) # 如果当前节点为空，也就意味着val找到了合适的位置，此时创建节点直接返回。
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if root.val < val:
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root.right = self.insertIntoBST(root.right, val) # 递归创建右子树
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if root.val > val:
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root.left = self.insertIntoBST(root.left, val) # 递归创建左子树
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return root
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```
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```scala []
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/**
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* Definition for a binary tree node.
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* class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
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* var value: Int = _value
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* var left: TreeNode = _left
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* var right: TreeNode = _right
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* }
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*/
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object Solution {
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def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
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if (root != null) {
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if (root.value < `val`) {
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root.right = insertIntoBST(root.right, `val`)
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}
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else {
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root.left = insertIntoBST(root.left, `val`)
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}
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} else {
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return new TreeNode(`val`)
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}
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root
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}
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}
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```
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```swift []
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class Solution {
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func insertIntoBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
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if root == nil{return TreeNode(val)}
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if val<root!.val{
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root?.left=insertIntoBST(root?.left, val)
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}else{
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root?.right=insertIntoBST(root?.right, val)
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}
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return root
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}
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}
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```

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