Construct Binary Tree from Inorder and Postorder Traversal with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nWe start by creating a hashmap called inorder_map to store the indices of
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nWe start by creating a hashmap called inorder_map to store the indices of each element in the inorder list. This will allow us to quickly find the index of a given value in constant time later on.\n\nThen, we define a recursive helper function called build that takes two arguments: the start and end indices of the current subtree we\'re working on. If the start index is greater than the end index, we\'ve reached the end of a branch and we should return None.\n\nInside the build function, we pop the last value from the postorder list and create a new node with that value. Then, we use the inorder_map to find the index of the node in the inorder list.\n\nSince we\'re working with the postorder traversal, we first recursively build the right subtree by calling build with the start index set to index + 1 and the end index set to end. Then, we recursively build the left subtree by calling build with the start index set to start and the end index set to index - 1.\n\nFinally, we call the build function with the start index set to 0 and the end index set to len(inorder) - 1, which will build the entire tree. The build function will return the root node of the tree.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:\n        # Use a hashmap to store the indices of each element in the inorder list\n        inorder_map = {}\n        for i, val in enumerate(inorder):\n            inorder_map[val] = i\n        \n        # Define a recursive helper function to build the tree\n        def build(start, end):\n            # Base case: the start index is greater than the end index\n            if start > end:\n                return None\n            \n            # Create a new node with the last value in the postorder list\n            val = postorder.pop()\n            node = TreeNode(val)\n            \n            # Find the index of the node in the inorder list\n            index = inorder_map[val]\n            \n            # Recursively build the right subtree first, since we\'re working with the postorder traversal\n            node.right = build(index + 1, end)\n            # Then build the left subtree\n            node.left = build(start, index - 1)\n            \n            return node\n        \n        # Call the helper function to build the tree\n        return build(0, len(inorder) - 1)\n\n```
['Array', 'Hash Table', 'Divide and Conquer', 'Python', 'Python3']