if we say the first node is a0, then inorder could be splited into 3 parts Al + [a0] + Ar. Al is the inorder of the left subtree of the node a0 while Ar is the right subtree.
Here comes this recursive solution.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if len(preorder) == 0:
return None
mid = inorder.index(preorder[0])
node = TreeNode(preorder[0])
node.left = self.buildTree(preorder[1:mid + 1], inorder[:mid])
node.right = self.buildTree(preorder[mid + 1:], inorder[mid + 1:])
return node