题解 | #重建二叉树#

# 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