# LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal

## Description

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

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```

## Explanation

find inorder root position and update preorder list

## Python 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:

return self.helper(preorder, inorder);

def helper(self, preorder, inorder):

if not inorder:
return None

inorder_position = inorder.index(preorder[0])

root = TreeNode(preorder[0])
root.left = self.helper(preorder[1 : 1 + inorder_position], inorder[0: inorder_position])
root.right = self.helper(preorder[inorder_position + 1:], inorder[inorder_position + 1:])

return root``````
• Time complexity: O(N).
• Space complexity: O(N).