701. Insert into a Binary Search Tree

Medium LeetCode

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, 10^4].
• -10^8 <= Node.val <= 10^8
• All the values Node.val are unique.
• -10^8 <= val <= 10^8
• It's guaranteed that val does not exist in the original BST.

How to solve the problem

Code

• Recursion (DFS) For BST

# 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) # We found the position to insert 'val'
        if val < root.val:
            root.left = self.insertIntoBST(root.left, val)
        else:
            root.right = self.insertIntoBST(root.right, val)
        return root

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)