# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None def constructTree(preOrder,vinOrder): # 递归退出条件 if len(preOrder) == 0: return None # 根节点 root_val = preOrder[0] root = TreeNode(root_val) index = vinOrder.index(root_val) # 左子树 leftnode = constructTree(preOrder[1:index+1], vinOrder[:index]) # 柚子树 rightnode = constructTree(preOrder[index+1:],vinOrder[index+1:]) root.left = leftnode root.right = rightnode return root class Solution: def reConstructBinaryTree(self , preOrder: List[int], vinOrder: List[int]) -> TreeNode: # write code here # 根据前序遍历确定中间节点，根据中序遍历，分左右子树 return constructTree(preOrder,vinOrder)