PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtre…

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 postorde…

1099. Build A Binary Search Tree (30) 时间限制 100 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 Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only…

链接:1135. Is It A Red-Black Tree (30) 红黑树的性质: (1) Every node is either red or black. (2) The root is black. (3) Every leaf (NULL) is black. (4) If a node is red, then both its children are black. (5) For each node, all simple paths from the node to de…

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. And it is a simple standard routine to print the numbers in level-order. H…

(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…

1135. Is It A Red-Black Tree (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties: (1) Every node is eith…

题意:输入一个正整数N(<=1000),接着输入N个非负整数(<=2000),输出完全二叉树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; int cnt; int n; void dfs(int x){ if(x>n) return; dfs(x*); tree[x]=a[++cnt]; dfs(x*+); }…

题意: 输入三个正整数N,M,S(N<=100,M<N,S<=2^30)分别代表数的结点个数,非叶子结点个数和需要查询的值,接下来输入N个正整数(<1000)代表每个结点的权重,接下来输入M行,每行包括一个两位数字组成的数代表非叶子结点的编号以及数字x表示它的孩子结点个数,接着输入x个数字表示孩子结点的编号.以非递增序输出从根到叶子结点的路径权重,它们的和等于S. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<…

题意: 输入一个正整数N(<=10000),接下来输入N个字符串,每个字符串包括至多8个字符,均为数字0~9.输出由这些字符串连接而成的最小数字(不输出前导零). trick: 数据点0只包含没有0的串. 数据点2,5,6包含最小串全部为0且次小串含有前导零的数据. 如果所有的数字均为0,输出一个0即可. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std;…