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 pre int整型一维数组 
# @param vin int整型一维数组 
# @return TreeNode类
#
class Solution:
    def reConstructBinaryTree(self , pre: List[int], vin: List[int]) -> TreeNode:
        # write code here
        def build(preStart:int, preEnd:int, vinStart:int, vinEnd:int):
            if preStart > preEnd:
                return None
            rootIndex = preStart
            vinroot = index[pre[rootIndex]]
            root = TreeNode(pre[rootIndex])
            leftSize = vinroot - vinStart
            root.left = build(preStart+1, preStart+leftSize, vinStart, vinroot-1)
            root.right = build(preStart+leftSize+1, preEnd, vinroot+1, vinEnd)

            return root

        index = {element:i for i, element in enumerate(vin)}
        n = len(pre)
        return build(0, n-1, 0, n-1)