Problem description

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Note 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.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3
And the value to insert: 5
You can return this binary search tree:

         4
       /   \
      2     7
     / \   /
    1   3 5
This tree is also valid:

         5
       /   \
      2     7
     / \   
    1   3
         \
          4

Solution

1. If root is None. We just create node with val and return that node
2. If root is not none and root.val < val, we need to insert into root’s right subtree and make root.right as the modified right subtree’s root
3. If root is not none and root.val > val, we need to insert into root’s left subtree and make root.left as the modified left subtree’s root.

Below is my python implementation

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
        if not root:
            return TreeNode(val)
        if root.val < val:
            right = self.insertIntoBST(root.right, val)
            root.right = right
        if root.val > val:
            left = self.insertIntoBST(root.left, val)
            root.left = left
        return root