Construct BST from preorder & inorder.md

Problem

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of 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.

optimized_solution.py

# Definition for a binary tree node.
# 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: List[int], inorder: List[int]) -> Optional[TreeNode]:
		# base case
		if not preorder or not inorder:
            return None

		preorder_queue = collections.deque(preorder)
		inorder_map = {value: key for key, value in enumerate(inorder)}

		def buildTreeHelper(left_index: int, right_index: int):
			
			# base case for indices
			if left_index > right_index:
				return None
            
			root_value = preorder_queue.popleft()  # keep track of value
			curr_root = TreeNode(val=root_value)
			midpoint = inorder_map[root_value]  # O(1) by using dict
			curr_root.left = buildTreeHelper(left_index, midpoint-1)
			curr_root.right = buildTreeHelper(midpoint+1, right_index)
			return curr_root

        return buildTreeHelper(0, len(preorder) - 1)  # only utilize the index of preorder, avoided time consumption creating lists for recursive calls

working_solution.py

# Definition for a binary tree node.
# 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: List[int], inorder: List[int]) -> Optional[TreeNode]:
        
		if not preorder or not inorder:
			return None
        
		curr_root = TreeNode(val=preorder[0])  # root is 1st element in preorder array
		midpoint = inorder.index(preorder[0])  # 1st recursive value: 1
        
		curr_root.left = self.buildTree(preorder=preorder[1:midpoint+1], inorder=inorder[:midpoint])  # construct left subtree
		curr_root.right = self.buildTree(preorder=preorder[midpoint+1:], inorder=inorder[midpoint+1:])  # construct right subtree
        
		return curr_root