Question Link:

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

The basic idea is same to that of Construct Binary Tree from Inorder and Postorder Traversal. We solve it using a recursive function. First, we find preorder[0] in inorder, let’s say inorder[i] == preorder, then construct the left tree from preorder[1..i] and inorder[0..i-1] and the right tree from preorder[i+1..n-1] and inorder[i+1..n-1].

The code is as follows.

[-] Python code accepted by LeetCode OJ 
# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param preorder, a list of integers
    # @param inorder, a list of integers
    # @return a tree node
    def buildTree(self, preorder, inorder):
        n = len(preorder)
        if n == 0:
            return None
        elif n == 1:
            return TreeNode(preorder[0])
        else:
            mid_inorder = inorder.index(preorder[0])
            root = TreeNode(preorder[0])
            root.left = self.buildTree(preorder[1:mid_inorder+1], inorder[:mid_inorder])
            root.right = self.buildTree(preorder[mid_inorder+1:], inorder[mid_inorder+1:])
            return root