Construct Binary Tree From Preorder And Inorder Traversal [Pythonn]

In this blog post, we will delve into a fascinating problem: Constructing a binary tree from its preorder and inorder traversals.

If you are a beginner or looking for a Python solution, you're in the right place.

We'll walk through the problem, provide a Python solution, explain the reasoning behind our approach, and analyze the time and space complexity.

Problem Overview

The problem at hand is to construct a binary tree given two integer arrays: preorder and inorder.

The preorder array represents the preorder traversal of the tree, while the inorder array represents the inorder traversal.

The task is to reconstruct and return the binary tree based on these two traversals.

Let's consider an example to understand this problem better:

Example:

Input:

preorder = [3, 9, 20, 15, 7]
inorder = [9, 3, 15, 20, 7]


Output:

[3, 9, 20, null, null, 15, 7]


In this example, we can see that the given preorder and inorder arrays do contain enough information to reconstruct the original binary tree.

Understanding the Constraints

Before diving into the solution, let's understand the constraints imposed by the problem:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• Both preorder and inorder consist of unique values.
• Each value in inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

These constraints help us realize that our solution needs to be efficient and handle a variety of input scenarios.

Construct Binary Tree From Preorder And Inorder Traversal LeetCode Problem Solution

Now, let's explore a Python solution to tackle this problem.

We'll follow a recursive approach to build the binary tree.

1. Bruteforce Approach

Before diving into the efficient solution, let's briefly discuss a brute-force approach.

We could potentially construct the binary tree by trying every combination of nodes.

However, this approach has exponential time complexity and is not practical for large inputs, which violate the constraints.

Therefore, we'll focus on the efficient approach.

2. An Efficient Approach with Pseudocode and Python Code

The efficient approach to construct the binary tree from the preorder and inorder traversals involves using recursion.

We already know two crucial facts:

1. The first value in the preorder traversal is always the root.
2. The inorder traversal divides the tree into left and right subtrees.

Here's a step-by-step approach with a Python code snippet:

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def buildTree(self, preorder, inorder):
        if not preorder or not inorder:
            return None

        # The first value in preorder is the root.

root = TreeNode(preorder[0])

        # Find the index of the root value in inorder.

mid = inorder.index(preorder[0])

        # Recursively construct the left and right subtrees.

root.left = self.buildTree(preorder[1:mid + 1], inorder[:mid])
        root.right = self.buildTree(preorder[mid + 1:], inorder[mid + 1:])

        return root

The code above defines a TreeNode class for constructing nodes in the binary tree.

The buildTree function recursively constructs the binary tree.

The base cases for recursion are:
– If preorder or inorder is empty, return None.
– If both preorder and inorder are non-empty, we follow these steps:
1. Create a TreeNode with the first value in preorder.
2. Find the index (mid) of the root value in inorder.
3. Recursively construct the left subtree using appropriate subarrays.
4. Recursively construct the right subtree using appropriate subarrays.

This efficient approach ensures that we can reconstruct the binary tree while adhering to the given constraints.

Time and Space Complexity

Let's analyze the time and space complexity of our solution.

• Time Complexity: The time complexity of our solution is O(N), where N is the number of nodes in the binary tree.

This is because we visit each node once.
– Space Complexity: The space complexity is O(N) as well, where N is the number of nodes in the binary tree.

This space is used for the recursive call stack.

Reasoning Behind Our Approach

The reasoning behind our approach lies in the characteristics of preorder and inorder traversals.

Preorder traversal always starts with the root, allowing us to identify the root node.

In contrast, inorder traversal separates the tree into left and right subtrees, indicating how many nodes belong to each side.

By leveraging these characteristics, we can recursively construct the binary tree with the given arrays.

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Conclusion

In this blog post, we've explored the LeetCode problem Construct Binary Tree From Preorder And Inorder Traversal We've provided a Python solution that efficiently constructs a binary tree based on the provided preorder and inorder traversals.

Our approach follows the characteristics of these traversals, allowing us to determine the root and divide the tree into left and right subtrees.

With a time complexity of O(N) and space complexity of O(N), this solution is capable of handling the constraints imposed by the problem.

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Feel free to comment, ask questions, make suggestions, and share this content with others to help them understand this problem better.

For the original question and more details, you can visit the LeetCode problem here.

Happy coding!