Day 7 of 100 days dsa coding challenge

Taking on a new challenge: solving GeeksforGeeks POTD daily and sharing my solutions! 💻🔥
The goal: sharpen problem-solving skills, level up coding, and learn something new every day. Follow my journey! 🚀

100DaysOfCode #CodingChallenge #ProblemSolving #GeeksforGeeks #DeveloperJourney

Problem:

https://www.geeksforgeeks.org/problems/construct-tree-from-preorder-postorder/1

Construct Tree from Preorder & Postorder

Difficulty: Medium Accuracy: 88.3%

Given two arrays pre[] and post[] that represent the preorder and postorder traversals of a full binary tree. Your task is to construct the binary tree and return its root.
Note: Full Binary Tree is a binary tree where every node has either 0 or 2 children. The preorder and postorder traversals contain unique values, and every value present in the preorder traversal is also found in the postorder traversal.
Examples:

Input: pre[] = [1, 2, 4, 8, 9, 5, 3, 6, 7], post[] = [8, 9, 4, 5, 2, 6, 7, 3, 1]
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9]
Explanation: The tree will look like

Input: pre[] = [1, 2, 4, 5, 3, 6, 7] , post[] = [4, 5, 2, 6, 7, 3, 1]
Output: [1, 2, 3, 4, 5, 6, 7]
Explanation: The tree will look like

Constraints:
1 ≤ number of nodes ≤ 103
1 ≤ pre[i], post[i] ≤ 104

Solution:
class Node:
def init(self, val):
self.data = val
self.left = None
self.right = None

class Solution:
def constructTree(self, pre, post):
self.index = 0
n = len(pre)
post_index = {val: i for i, val in enumerate(post)}

    def build(l, r):
        if self.index >= n or l > r:
            return None
        root = Node(pre[self.index])
        self.index += 1
        if l == r or self.index >= n:
            return root
        mid = post_index[pre[self.index]]
        root.left = build(l, mid)
        root.right = build(mid + 1, r - 1)
        return root

    return build(0, n - 1)