题目 Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the posto…

题目 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one t…

No.1119 题目:由前序后序二叉树序列,推中序,判断是否唯一后输出一组中序序列 思路:前序从前向后找,后序从后向前找,观察正反样例可知,前后序树不唯一在于单一子树是否为左右子树. 判断特征:通过查找后序序列中最后一个结点的前一个在先序中的位置,来确定是否可以划分左右孩子,如果不能,  就将其划分为右孩子(或左孩子),递归建树. 中序遍历输出. #include <iostream> using namespace std; const int maxn = 31; int n, index…

[题目链接] [题意] 根据二叉树的前序和后序序列,如果中序序列唯一,输出Yes,如果不唯一输出No,并输出这个中序序列. [题解] 众所周知,二叉树是不能够根据前序和中序建立的,为什么呢?首先需要明确先序序列的遍历顺序是:根左右,后序序列的遍历顺序是:左右根. 然后我们来说一下这个样例(为了更好的说明不唯一性,没有用题目给出的样例): 前序:1 4 6 3 2 5 7 后序:3 6 4 5 7 2 1 首先1肯定是整棵二叉树的根节点,然后根据前序序列我们可以知道4是1的子树(此时还不确定是左子…

1020 Tree Traversals (25 分)   Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tr…

题目 Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree. Input Specification: Ea…

题目 Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree,…

(先说一句,题目还不错,很值得动手思考并且去实现.) 题意:根据前序遍历和后序遍历建树,输出中序遍历序列,序列可能不唯一,输出其中一个即可. 已知前序遍历和后序遍历序列,是无法确定一棵二叉树的,原因在于如果只有一棵子树可能是左孩子也有可能是右孩子.由于只要输出其中一个方案,所以假定为左孩子即可.下面就是如何根据前序和后序划分出根节点和左右孩子,这里需要定义前序和后序的区间范围,分别为[preL,preR],[postL,postR]. 一开始区间都为[1,n],可以发现前序的第一个和后序的最后一…

1021. Deepest Root (25) 时间限制 1500 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root t…

//题目 通过后续遍历 中序遍历 得出一棵树 ,然后按树的层次遍历打印 PS:以前对于这种用指针的题目是比较头痛的,现在做了一些链表操作后,感觉也不难 先通过后续中序建一棵树,然后通过BFS遍历这棵树 提供测试样例 44 1 3 22 3 1 41010 7 6 9 8 5 4 1 3 27 10 6 2 5 9 8 3 1 4 //题目 通过后续遍历 中序遍历 得出一棵树 ,然后按树的层次遍历打印 #include<stdio.h> #include<iostream> #inc…