Construct Binary Tree from Preorder and Inorder Traversal.md

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7
  

My code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque
class Solution:
   def buildTree(self, preorder, inorder):
       """
       :type preorder: List[int]
       :type inorder: List[int]
       :rtype: TreeNode
       """
       return self.tree(deque(preorder), inorder)
   def tree(self,preorder, inorder):
       if not inorder:
           return None
       root_val=preorder.popleft()
       root=TreeNode(root_val)
       index=inorder.index(root_val)
       root.left=self.tree(preorder,inorder[:index])
       root.right=self.tree(preorder,inorder[index+1:])
       return root
       

Solution: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/