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:
Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:
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 will be in the range
[0, 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.
SOLUTION:
# 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:
if val > root.val:
root.right = self.insertIntoBST(root.right, val)
else:
root.left = self.insertIntoBST(root.left, val)
return root
else:
return TreeNode(val=val)
Top comments (0)