# -*- coding:utf-8 -*- # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # 返回构造的TreeNode根节点 def reConstructBinaryTree(self, pre, tin): # write code here if len(pre) == 0: return None if len(pre) == 1: return TreeNode(pre[0]) else: flag = TreeNode(pre[0]) flag.left = self.reConstructBinaryTree(pre[1:tin.index(pre[0])+1],tin[:tin.index(pre[0])]) flag.right = self.reConstructBinaryTree(pre[tin.index(pre[0])+1:],tin[tin.index(pre[0])+1:] ) return flag分情况考虑,无节点,单节点除外,剩下的,按照前序遍历和中序遍历的特点加进去。
笔试算法---二叉树重建
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转载自blog.csdn.net/mr_ming_/article/details/79438885
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