105 Construct Binary Tree From Preorder And Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

 

Example 1:

**Input:** preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
**Output:** [3,9,20,null,null,15,7]

Example 2:

**Input:** preorder = [-1], inorder = [-1]
**Output:** [-1]

 

Constraints:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder and inorder consist of **unique** values.
Each value of inorder also appears in preorder.
preorder is **guaranteed** to be the preorder traversal of the tree.
inorder is **guaranteed** to be the inorder traversal of the tree.

# 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]:
        
        if len(inorder) == 0 and len(preorder) == 0:
            return None
        
        root = TreeNode(preorder[0])
        inorder_index = 0

        inorder_index = inorder.index(preorder[0])

        left_inorder = inorder[:inorder_index]
        right_inorder = inorder[inorder_index+1:]
       
        left_preorder = preorder[1:inorder_index+1]
        right_preorder = preorder[inorder_index+1:]
        
        root.left = self.buildTree( left_preorder, left_inorder)
        root.right = self.buildTree(right_preorder, right_inorder)
        return root