105. Construct Binary Tree from Preorder and Inorder Traversal
Medium
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:
Example 2:
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.
Solution
# 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 not preorder or not inorder:
return None
root = preorder[0]
idx = inorder.index(root)
n = len(inorder[:idx]) + 1
return TreeNode(
root,
self.buildTree(preorder[1:n], inorder[:idx]),
self.buildTree(preorder[n:], inorder[idx + 1 :]),
)