题解 | #重建二叉树#

思路：先序遍历首位为根，然后基于根在中序遍历中拆分左、右子树，依次递归构建

注意：

1、root：1

2、idx：vin[root.val]

3、len(left) = [:idx]

4、len(right) = [idx+1:]

class Solution:
    def reConstructBinaryTree(self , pre: List[int], vin: List[int]) -> TreeNode:
        
        if len(pre) == 0 and len(vin) == 0:
            return
        
        root = TreeNode(pre[0])
        idx = vin.index(root.val)
        
        left = self.reConstructBinaryTree(pre[1:1+idx], vin[:idx])
        right = self.reConstructBinaryTree(pre[1+idx:], vin[idx+1:])
        root.left = left
        root.right = right
        
        return root