Build Tree From Preorder and Postorder given Inorder [LeetCode 105/106]

FromPostOrder.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, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        
        root = None
        
        if inorder and postorder:
            
            # Get the current root we are looking for
            val = postorder[-1]
            
            # Find the index of the root in inorder list, this is the same index
            # in the postorder to divide the list
            idx = inorder.index(val)
            
            # Make root with the current value 
            root = TreeNode(val)
            
            # Make left subtree, take inorder upto index and postorder upto index
            root.left = self.buildTree(inorder[:idx], postorder[:idx])
            
            # Make right subtree, take inorder from the next index and postorder
            # from the current index to the second last index
            root.right = self.buildTree(inorder[idx+1:], postorder[idx:-1])
            
        return root

FromPreOrder.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]:
        
        root = None
        
        if inorder and preorder:
            
            # Get the current root we are looking for
            val = preorder[0]
            
            # Find the index of the root in inorder list, this is the same index
            # in the preorder to divide the list
            idx = inorder.index(val)
            
            # Make root with the current value 
            root = TreeNode(val)
            
            # Make left subtree, take inorder upto index and preorder from the first
            # upto the current index
            root.left = self.buildTree(preorder[1:idx+1], inorder[:idx])
            
            # Make right subtree, take inorder from the next index and preorder
            # from the next index
            root.right = self.buildTree(preorder[idx+1:], inorder[idx+1:])
            
        return root