Problem of The Day: Insert into a Binary Search Tree

Problem Statement

leetcode problem link

Brute Force [Accepted]

# 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]:
        def dfs(node, parent):
            if not node:
                if parent.val > val:
                    parent.left = TreeNode(val)
                else:
                    parent.right = TreeNode(val)
                return

            if val > node.val:
                dfs(node.right, node)
            else:
                dfs(node.left, node)

        if not root:
            return TreeNode(val)

        dfs(root, val)

        return root

Editorial

Approach 1: Recursion

class Solution:
    def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
        if not root:
            return TreeNode(val)

        if val > root.val:
            # insert into the right subtree
            root.right = self.insertIntoBST(root.right, val)
        else:
            # insert into the left subtree
            root.left = self.insertIntoBST(root.left, val)
        return root

Approach 2: Iteration

class Solution:
    def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
        node = root
        while node:
            # insert into the right subtree
            if val > node.val:
                # insert right now
                if not node.right:
                    node.right = TreeNode(val)
                    return root
                else:
                    node = node.right
            # insert into the left subtree
            else:
                # insert right now
                if not node.left:
                    node.left = TreeNode(val)
                    return root
                else:
                    node = node.left
        return TreeNode(val)