What is the time complexity of Construct Binary Search Tree from Preorder Traversal?

The optimal solution for Construct Binary Search Tree from Preorder Traversal runs in O(n) time and uses O(n) space. The time complexity is O(n) since we process each element exactly once. The space complexity is O(n) due to the recursive call stack in the worst case.

What is the brute force solution for Construct Binary Search Tree from Preorder Traversal?

The brute force approach for Construct Binary Search Tree from Preorder Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves inserting each value from the preorder array into the binary search tree one by one. This method is straightforward but inefficient since each insertion may require traversing the tree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Construct Binary Search Tree from Preorder Traversal?

Construct Binary Search Tree from Preorder Traversal uses the following concepts: Array, Stack, Tree, Binary Search Tree, Monotonic Stack, Binary Tree. The recommended patterns to study are: Recursion, Tree Traversals.

What are the interview tips for Construct Binary Search Tree from Preorder Traversal?

Clarify the properties of a BST and how preorder traversal works before starting. Think about edge cases, such as the smallest input size. Practice building trees from traversals to become familiar with the process.

What are common mistakes when solving Construct Binary Search Tree from Preorder Traversal?

Not maintaining the bounds correctly, leading to incorrect tree structure. Confusing the order of insertion in a BST, especially when handling left and right children.

#1008

Construct Binary Search Tree from Preorder Traversal

Medium

ArrayStackTreeBinary Search TreeMonotonic StackBinary TreeRecursionTree Traversals

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses the properties of the preorder traversal and recursion to build the BST efficiently. By maintaining a range for valid values, we can decide where to place each node without needing to traverse the tree repeatedly.

⚙️

Algorithm

3 steps

1Step 1: Define a recursive function that takes the current bounds (lower and upper) for valid node values.
2Step 2: If the current value is within bounds, create a new node and recursively build the left and right subtrees with updated bounds.
3Step 3: Return the constructed node.

solution.py22 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 bst_from_preorder(preorder):
9    def helper(lower, upper):
10        nonlocal index
11        if index == len(preorder):
12            return None
13        val = preorder[index]
14        if val < lower or val > upper:
15            return None
16        index += 1
17        root = TreeNode(val)
18        root.left = helper(lower, val)
19        root.right = helper(val, upper)
20        return root
21    index = 0
22    return helper(float('-inf'), float('inf'))

ℹ

Complexity note: The time complexity is O(n) since we process each element exactly once. The space complexity is O(n) due to the recursive call stack in the worst case.

1The preorder traversal gives us the root first, allowing us to build the tree recursively.
2Maintaining bounds for valid node values helps in deciding where to place each node efficiently.

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