1115 Counting Nodes in a BST(30 分) 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 or equal to the node's key. The right subtree…

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 or equal to the node's key. The right subtree of a node contains only nodes with…

1115 Counting Nodes in a BST (30 分) 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 or equal to the node's key. The right subtre…

1115 Counting Nodes in a BST (30 分) 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 or equal to the node's key. The right subtre…

1115. Counting Nodes in a BST (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 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 nod…

题意:给出一个序列,构建二叉搜索树(BST),输出二叉搜索树最后两层的节点个数n1和n2,以及他们的和sum: n1 + n2 = sum 递归建树,然后再dfs求出最大层数,接着再dfs计算出最后两层的节点个数,也可以直接一遍dfs,顺便存储各个层的节点数. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #define LEFT 1 #defin…

https://pintia.cn/problem-sets/994805342720868352/problems/994805355987451904 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 or…

题目 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The lef subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes wi…

题目:二叉排序树,统计最后两层节点个数 思路:数组格式存储,insert建树,dfs遍历 #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const int INF = 0x7FFFFFFF; const int maxn = 1e5 + 10; int n, cnt[max…

简单题.统计一下即可. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<algorithm> using namespace std; +; struct Node { int left; int right; in…

题意: 输入一个正整数N(<=1000),接着输入N个整数([-1000,1000]),依次插入一棵初始为空的二叉排序树.输出最底层和最底层上一层的结点个数之和,例如x+y=x+y. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct Node{ int value; int vis; int level; int visl,…

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 or equal to the node's key. The right subtree of a node contains only nodes with…

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 or equal to the node's key. The right subtree of a node contains only nodes with…

题意:给出一棵二叉搜索树的插入序列,要求该树最后两层的结点个数. 思路:在树结点中增加一个数据域layer,表示该结点所在的层次.另外,设置数组level[]和变量maxLevel,level[i]表示第i层的结点个数,maxLevel表示树的最大层次,在层序遍历时更新即可. 代码: #include <cstdio> #include <queue> using namespace std; ]={}; ; struct Node{ int val; int layer; Nod…

1115. Counting Nodes in a BST (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 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 nod…

Source: PAT A1115 Counting Nodes in a BST (30 分) Description: 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 or equal to the no…

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 or equal to the node's key. The right subtree of a node contains only nodes with…

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 or equal to the node's key. The right subtree of a node contains only nodes with…

题目一:给定一个数组,升序数组,将他构建成一个BST 思路:升序数组,这就类似于中序遍历二叉树得出的数组,那么根节点就是在数组中间位置,找到中间位置构建根节点,然后中间位置的左右两侧是根节点的左右子树,递归的对左右子树进行处理,得出一颗BST 代码: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val…

1004 Counting Leaves (30分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0…

A family hierarchy is usually presented by a pedigree tree.  Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tre…

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a…

1049 Counting Ones (30)(30 分) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12.…

1049 Counting Ones (30 分) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Inp…

看别人的题解懂了一些些    参考<编程之美>P132 页<1 的数目> #include<iostream> #include<stdio.h> using namespace std; int getone(int n) { int ans=0,base=1,right,left,now; while(n/base) { right=n%base; left=n/(base*10); now=(n/base)%10; if(now==0)ans+=lef…

要统计1到N之间‘1’的个数,如数11包含2个1.所以当N=12时,答案为5. 思想: 找规律,假设ans[N]表示1到N的‘1’的个数,则有a[100]=(a[10]-1)*9+10+a[10]-1+1; 先打表求出1ek的答案: 然后对N由高到低逐位拆分. 有种情况要特别注意: 当N=100001时,高位出现1时要累加到后面第一个非0位数上. #include<iostream> #include<cstring> #include<cstdio> #include…

［抄题］: Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST. Basically, the deletion can be divided into two stages: Search for a node to remove. If…

［抄题］: Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements. Example 1: Input: root = [3,1,4,null,2], k = 1 3 / \ 1 4 \   2 Output: 1 Exam…

这道题学到了东西. /* 一开始想着中序遍历,但是解码的时候才发现,中序遍历并不能唯一得确定二叉树. 后来看了网上的答案,发现先序遍历是可以的,观察了一下,对于BST,先序遍历确实是可以 唯一得确定. 对于一般的二叉树,必须是前序和中序或者后序和中序才能唯一确定,其中中序的作用就是 确定中心位置,以确定左右子树的前序(后序)的范围 但是对于BST根据性质,左子树的前序序列最大的就是1st节点,所以找到序列中第一个比 1st节点大的节点就是右子树的开始,这样就不需要中序了 还有一种方法,保存二叉树…

较为简单.小于大于的都走一遍就可以AC了 #include <stdio.h> #include <memory.h> #include <math.h> #include <string> #include <vector> #include <set> #include <stack> #include <queue> #include <algorithm> #include <map&…