# 701 Insert into a Binary Search Tree

You are given the `root` node of a binary search tree (BST) and a `value` to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Example 1:

``````      4             4
/ \         /    \
2   7       2      7
/  \         / \    /
1    3       1   3  5

Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:
5
/ \
2   7
/ \
1   3
\
4
``````

Example 2:

``````Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]
``````

Example 3:

``````Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]
``````

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -108 <= `Node.val` <= 108
• All the values `Node.val` are unique.
• -108 <= `val` <= 108
• It's guaranteed that `val` does not exist in the original BST.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 `````` ``````# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: if not root: return TreeNode(val) if val > root.val: root.right = self.insertIntoBST(root.right, val) else: root.left = self.insertIntoBST(root.left, val) return root``````