PAT甲級
二叉樹的遍歷
樹的遍歷
- 1079. Total Sales of Supply Chain
- 1090. Highest Price in Supply Chain
- 1094. The Largest Generation
- 1004. Counting Leaves
- 1053. Path of Equal Weight (30)
BST
/*1043.先根據(jù)序列構(gòu)造出BST or mirror BST，然后再比較先序遍歷序列是否相等，然后再輸出后序遍歷的序列*/
#include<iostream>
#include<vector>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x) :val(x), left(NULL), right(NULL) {};
};
void insert(TreeNode* &root,int x)
{
if (root == NULL) {
root = new TreeNode(x);
return;
}
if (x < root->val) {
insert(root->left, x);
}
if (x >= root->val) {
insert(root->right, x);
}
}
void mirrorinsert(TreeNode* &root, int x)
{
if (root == NULL) {
TreeNode* node = new TreeNode(x);
root = node;
return;
}
if (x >= root->val) {
mirrorinsert(root->left, x);
}
if (x < root->val) {
mirrorinsert(root->right, x);
}
}
TreeNode* buildTree(vector<int> &input)
{
int Size = input.size();
TreeNode* root = new TreeNode(input[0]);
if (Size == 1)
return root;
bool flag = true;
if (input[1] > input[0]) {
flag = false;//mirror bst
}
if (flag)//insert
{
for (int i = 1; i < Size; i++)
{
insert(root, input[i]);
}
}
else
{
for (int i = 1; i < Size; i++) {
mirrorinsert(root, input[i]);
}
}
return root;
}
vector<int> preorder(TreeNode* root)
{
TreeNode* cur = root;
vector<int> preOrder,tmp1,tmp2;
preOrder.push_back(cur->val);
if (root->left) {
tmp1 = preorder(root->left);
preOrder.insert(preOrder.end(), tmp1.begin(), tmp1.end());
}
if (root->right) {
tmp2 = preorder(root->right);
preOrder.insert(preOrder.end(), tmp2.begin(), tmp2.end());
}
return preOrder;
}
vector<int> postorder(TreeNode* root)
{
TreeNode* cur = root;
vector<int> postOrder, tmp1, tmp2;
if (!root)
return postOrder;
tmp1 = postorder(root->left);
postOrder.insert(postOrder.end(), tmp1.begin(), tmp1.end());
tmp2 = postorder(root->right);
postOrder.insert(postOrder.end(), tmp2.begin(), tmp2.end());
postOrder.push_back(cur->val);
return postOrder;
}
int main()
{
int m;//樹的結(jié)點數(shù)
cin >> m;
vector<int> input(m);
for (int i = 0; i < m; i++) {
cin>>input[i];
}
TreeNode* tree = buildTree(input);
bool judge = true;
vector<int> pre = preorder(tree);
for (int i = 0; i < m; i++) {
if (pre[i] != input[i]) {
judge = false;
break;
}
}
if (judge) {
cout << "Yes" << endl;
vector<int> post = postorder(tree);
for (int i = 0; i < m-1; i++) {
cout << post[i] << " ";
}
cout << post[m - 1];
}
else {
cout << "No" << endl;
}
return 0;
}
/*1064:給定一組序列，構(gòu)造完全二叉搜索樹，輸出層序遍歷序列*/
/*思路：完全二叉樹用數(shù)組來表示，中序遍歷這個數(shù)組，按遍歷的順序把數(shù)字從小到大賦值即構(gòu)造出完全二叉搜索樹
比如，第一次訪問下標(biāo)為7的數(shù)組a節(jié)點，則賦值a[7] = min*/
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int i=0;//表示排好序的數(shù)組下標(biāo)
void buildCBST(vector<int> &cbst,vector<int> &input,int index) {//遞歸
if (index >= input.size()) {
return;
}
buildCBST(cbst, input, index * 2 + 1);//左子樹
cbst[index] = input[i++];
buildCBST(cbst,input,index*2+2);//右子樹
}
int main()
{
int n;
cin >> n;
vector<int> input(n),cbst(n);
for (int j = 0; j < n; j++) {
cin >> input[j];
}
sort(input.begin(), input.end());
/*for (int j = 0; j < n; j++) {
cout << input[j]<<" ";
}*/
buildCBST(cbst, input, 0);
for (int j = 0; j < cbst.size()-1; j++) {
cout<< cbst[j]<<" ";
}
cout << cbst[n - 1];
return 0;
}
/*1107. Social Clusters */
#pragma warning(disable:4996)
#include<cstdio>
#include<algorithm>
const int N = 1001;
using namespace std;
int id[N] = { 0 };
int course[N] = { 0 };
int isroot[N] = { 0 };
void init(int n) {
for (int i = 1; i <=n; i++) {
id[i] = i;
isroot[i] = 0;
}
}
int root(int x) {
if (x == id[x])return x;
else {
int r = root(id[x]);
id[x] = r;
return r;
}
}
bool find(int p, int q) {
return root(p) == root(q);
}
void unit(int q, int p) {
if (!find(q, p)) {//p,q不在一個集合里,把q集合加入p集合
id[root(q)] = id[root(p)];//q集合根節(jié)點指向p集合根節(jié)點
}
}
bool cmp(int a, int b) {
return a > b;
}
int main() {
int m,k,h;
scanf("%d\n", &m);
init(m);
for (int i = 1; i <= m; i++) {
scanf("%d:", &k);
for (int j = 0; j < k; j++) {
scanf("%d",&h);
if (course[h] == 0) {
course[h] = i;//第一個i選擇興趣h的作為根節(jié)點，以后也選擇這個興趣的和根節(jié)點加入一個集合
}
unit(i, root(course[h]));//實際上是i和root(一樣興趣的人）unit
}
}
for (int i = 1; i <= m; i++) {
isroot[root(i)]++;//根節(jié)點連著幾個人
}
int sum = 0;//幾個集合
for (int i = 1; i <= m; i++) {
if (isroot[i] != 0) {
sum++;
}
}
printf("%d\n",sum);
sort(isroot+1, isroot + m + 1, cmp);
for (int i = 1; i <sum; i++) {
printf("%d ", isroot[i]);
}
printf("%d", isroot[sum]);
return 0;
}
