Construct binary tree from preorder and inorder traversal
# 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 # Leaf node
root_val = preorder[0]
root = TreeNode(root_val)
idx = inorder.index(root_val)
in_left = inorder[:idx]
in_right = inorder[idx + 1:]
pre_left = preorder[1:len(in_left) + 1]
pre_right = preorder[len(in_left) + 1:]
root.left = self.buildTree(pre_left, in_left)
root.right = self.buildTree(pre_right, in_right)
return root
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 <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorderandinorderconsist of unique values.- Each value of
inorderalso appears inpreorder. preorderis guaranteed to be the preorder traversal of the tree.inorderis guaranteed to be the inorder traversal of the tree.