1119. Pre- and Post-order Traversals 引用

本文探讨了一种有效的二叉树序列化算法,并通过实例分析了左右子树划分的重要性。针对特定二叉树结构,文章指出常见的序列化误区及正确的划分方法。

直接看大神的算法吧,不能再棒!

http://blog.youkuaiyun.com/liuchuo/article/details/52505179

…………………………………………更新线………………………………………………

这里最重要的还是左右子树划分的问题。

第二次做这个题有点想当然的认为,先序NLR后续LRN,一定都是先序第一个等于后续最后一个然后确认节点和对应的L,R;

后面自己写了个树就发现

              1

       2      3

  4      5      6       7

先序:1 2 4 5 3 6 7

后序:4 5 2 6 7 3 1

再完成1的节点排序后,后面就无法进行上面的算法了。

2 4 5 6 7

5 2 6 7 3

所以还是要理解二叉树左右子树划分的方法,这样才能更好的去处理二叉树的问题。

 

American Heritage 题目描述 Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear tree in-order" and tree pre-order" notations. Your job is to create the `tree post-order" notation of a cow"s heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes. Here is a graphical representation of the tree used in the sample input and output: C / \ / \ B G / \ / A D H / \ E F The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree. The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree. The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root. ---------------------------------------------------------------------------------------------------------------------------- 题目大意: 给出一棵二叉树的中序遍历 (inorder) 和前序遍历 (preorder),求它的后序遍历 (postorder)。 输入描述 Line 1: The in-order representation of a tree. Line 2: The pre-o rder representation of that same tree. Only uppercase letter A-Z will appear in the input. You will get at least 1 and at most 26 nodes in the tree. 输出描述 A single line with the post-order representation of the tree. 样例输入 Copy to Clipboard ABEDFCHG CBADEFGH 样例输出 Copy to Clipboard AEFDBHGC c语言,代码不要有注释
06-16
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值