题解 | #重建二叉树#
重建二叉树
https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param preOrder int整型一维数组 # @param vinOrder int整型一维数组 # @return TreeNode类 # class Solution: def reConstructBinaryTree(self , preOrder: List[int], vinOrder: List[int]) -> TreeNode: # write code here if not preOrder or not vinOrder: return None root = TreeNode(preOrder[0]) rootindex = vinOrder.index(root.val) root.left = self.reConstructBinaryTree(preOrder[1:rootindex+1], vinOrder[:rootindex]) root.right = self.reConstructBinaryTree(preOrder[rootindex+1:], vinOrder[rootindex+1:]) return root