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

<p><strong>Notice</strong>&nbsp;that there may exist&nbsp;multiple valid ways for the&nbsp;insertion, as long as the tree remains a BST after insertion. You can return <strong>any of them</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/insertbst.jpg" style="width: 752px; height: 221px;" />
<pre>
<strong>Input:</strong> root = [4,2,7,1,3], val = 5
<strong>Output:</strong> [4,2,7,1,3,5]
<strong>Explanation:</strong> Another accepted tree is:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/bst.jpg" style="width: 352px; height: 301px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [40,20,60,10,30,50,70], val = 25
<strong>Output:</strong> [40,20,60,10,30,50,70,null,null,25]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
<strong>Output:</strong> [4,2,7,1,3,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in&nbsp;the tree will be in the range <code>[0,&nbsp;10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>8</sup> &lt;= Node.val &lt;= 10<sup>8</sup></code></li>
	<li>All the values <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>-10<sup>8</sup> &lt;= val &lt;= 10<sup>8</sup></code></li>
	<li>It&#39;s <strong>guaranteed</strong> that <code>val</code> does not exist in the original BST.</li>
</ul>

# 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 root is None:
            return TreeNode(val)

        res=root
        while root:
            if val<root.val:
                if root.left:
                    root=root.left
                else:
                    root.left=TreeNode(val)
                    break
            else:
                if root.right:
                    root=root.right
                else:
                    root.right=TreeNode(val)
                    break
        return res

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