How do you solve Insert into a Binary Search Tree on LeetCode?

Insert into a Binary Search Tree (LeetCode #701) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the properties of BSTs allows for efficient insertion.

What is the brute force solution for Insert into a Binary Search Tree?

The brute force approach for Insert into a Binary Search Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing the entire tree to find the correct position for the new value. This is straightforward but inefficient as it may require checking every node. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Insert into a Binary Search Tree?

Insert into a Binary Search Tree uses the following concepts: Tree, Binary Search Tree, Binary Tree. The recommended patterns to study are: Binary Search Tree, Tree Traversal.

What are the interview tips for Insert into a Binary Search Tree?

Explain your thought process clearly while coding. Consider edge cases and test your code with them. Practice writing both recursive and iterative solutions.

What are common mistakes when solving Insert into a Binary Search Tree?

Not handling edge cases like inserting into an empty tree. Confusing left and right child conditions during traversal.

#701

Insert into a Binary Search Tree

Medium

TreeBinary Search TreeBinary TreeBinary Search TreeTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses the properties of a BST to directly find the correct position for the new value without unnecessary traversals. This ensures we maintain the BST structure efficiently.

⚙️

Algorithm

5 steps

1Step 1: Start from the root of the BST.
2Step 2: Compare the new value with the current node's value.
3Step 3: If the new value is less, move to the left child; if greater, move to the right child.
4Step 4: Continue until you find a null position where the new value can be inserted.
5Step 5: Insert the new value at the found null position.

solution.py25 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def insertIntoBST(root: TreeNode, val: int) -> TreeNode:
9    if not root:
10        return TreeNode(val)
11    current = root
12    while True:
13        if val < current.val:
14            if current.left:
15                current = current.left
16            else:
17                current.left = TreeNode(val)
18                break
19        else:
20            if current.right:
21                current = current.right
22            else:
23                current.right = TreeNode(val)
24                break
25    return root

ℹ

Complexity note: The time complexity is O(n) in the worst case for a skewed tree, but the space complexity is O(1) since we are using iterative traversal without additional data structures.

1Understanding the properties of BSTs allows for efficient insertion.
2Iterative approaches can reduce the risk of stack overflow in deep trees.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.