105. 从前序与中序遍历序列构造二叉树 - 力扣（LeetCode）

# 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:  # 如果任一遍历为空，返回 Nonereturn None# 根节点是前序遍历的第一个元素root_val = preorder[0]root = TreeNode(root_val)# 找到根节点在中序遍历中的位置root_index = inorder.index(root_val)# 划分左子树和右子树left_inorder = inorder[:root_index]right_inorder = inorder[root_index + 1:]left_preorder = preorder[1:1 + len(left_inorder)]right_preorder = preorder[1 + len(left_inorder):]# 递归构建左子树和右子树root.left = self.buildTree(left_preorder, left_inorder)root.right = self.buildTree(right_preorder, right_inorder)return root