原文:
给定一个只包含小写字符的字符串字符串。问题是使用二叉树 打印字符及其出现频率。示例:
输入:str = " aaaaabannccccz " 输出:【a4b2n2 c4z】 解释:
输入:输出: g2e4k2s2for
进场:
- 从字符串中的第一个字符开始。
- 在二叉树中执行字符的级别顺序插入
- 选择下一个字符:
- 如果已经看到该字符，并且我们在级别顺序插入期间遇到它，则增加节点的计数。
- 如果到目前为止还没有看到该角色，请转到第 2 步。
- 对字符串的所有字符重复该过程。
- 打印应该输出所需输出的树的层次顺序遍历。
以下是上述方法的实现:
c
// c implementation of
// the above approach
#include
using namespace std;
// node in the tree where
// data holds the character
// of the string and cnt
// holds the frequency
struct node {
char data;
int cnt;
node *left, *right;
};
// function to add a new
// node to the binary tree
node* add(char data)
{
// create a new node and
// populate its data part,
// set cnt as 1 and left
// and right children as null
node* newnode = new node;
newnode->data = data;
newnode->cnt = 1;
newnode->left = newnode->right = null;
return newnode;
}
// function to add a node
// to the binary tree in
// level order
node* addinlvlorder(node* root, char data)
{
if (root == null) {
return add(data);
}
// use the queue data structure
// for level order insertion
// and push the root of tree to queue
queue q;
q.push(root);
while (!q.empty()) {
node* temp = q.front();
q.pop();
// if the character to be
// inserted is present,
// update the cnt
if (temp->data == data) {
temp->cnt ;
break;
}
// if the left child is
// empty add a new node
// as the left child
if (temp->left == null) {
temp->left = add(data);
break;
}
else {
// if the character is present
// as a left child, update the
// cnt and exit the loop
if (temp->left->data == data) {
temp->left->cnt ;
break;
}
// add the left child to
// the queue for further
// processing
q.push(temp->left);
}
// if the right child is empty,
// add a new node to the right
if (temp->right == null) {
temp->right = add(data);
break;
}
else {
// if the character is present
// as a right child, update the
// cnt and exit the loop
if (temp->right->data == data) {
temp->right->cnt ;
break;
}
// add the right child to
// the queue for further
// processing
q.push(temp->right);
}
}
return root;
}
// function to print the
// level order traversal of
// the binary tree
void printlvlorder(node* root)
{
// add the root to the queue
queue q;
q.push(root);
while (!q.empty()) {
node* temp = q.front();
// if the cnt of the character
// is more then one, display cnt
if (temp->cnt > 1) {
cout << temp->data << temp->cnt;
}
// if the cnt of character
// is one, display character only
else {
cout << temp->data;
}
q.pop();
// add the left child to
// the queue for further
// processing
if (temp->left != null) {
q.push(temp->left);
}
// add the right child to
// the queue for further
// processing
if (temp->right != null) {
q.push(temp->right);
}
}
}
// driver code
int main()
{
string s = "geeksforgeeks";
node* root = null;
// add individual characters
// to the string one by one
// in level order
for (int i = 0; i < s.size(); i ) {
root = addinlvlorder(root, s[i]);
}
// print the level order
// of the constructed
// binary tree
printlvlorder(root);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of
// the above approach
import java.util.*;
class gfg
{
// node in the tree where
// data holds the character
// of the string and cnt
// holds the frequency
static class node
{
char data;
int cnt;
node left, right;
};
// function to add a new
// node to the binary tree
static node add(char data)
{
// create a new node and
// populate its data part,
// set cnt as 1 and left
// and right children as null
node newnode = new node();
newnode.data = data;
newnode.cnt = 1;
newnode.left = newnode.right = null;
return newnode;
}
// function to add a node
// to the binary tree in
// level order
static node addinlvlorder(node root, char data)
{
if (root == null)
{
return add(data);
}
// use the queue data structure
// for level order insertion
// and push the root of tree to queue
queue q = new linkedlist();
q.add(root);
while (!q.isempty())
{
node temp = q.peek();
q.remove();
// if the character to be
// inserted is present,
// update the cnt
if (temp.data == data)
{
temp.cnt ;
break;
}
// if the left child is
// empty add a new node
// as the left child
if (temp.left == null)
{
temp.left = add(data);
break;
}
else
{
// if the character is present
// as a left child, update the
// cnt and exit the loop
if (temp.left.data == data)
{
temp.left.cnt ;
break;
}
// add the left child to
// the queue for further
// processing
q.add(temp.left);
}
// if the right child is empty,
// add a new node to the right
if (temp.right == null)
{
temp.right = add(data);
break;
}
else
{
// if the character is present
// as a right child, update the
// cnt and exit the loop
if (temp.right.data == data)
{
temp.right.cnt ;
break;
}
// add the right child to
// the queue for further
// processing
q.add(temp.right);
}
}
return root;
}
// function to print the
// level order traversal of
// the binary tree
static void printlvlorder(node root)
{
// add the root to the queue
queue q = new linkedlist();
q.add(root);
while (!q.isempty())
{
node temp = q.peek();
// if the cnt of the character
// is more then one, display cnt
if (temp.cnt > 1)
{
system.out.print((temp.data "" temp.cnt));
}
// if the cnt of character
// is one, display character only
else
{
system.out.print((char)temp.data);
}
q.remove();
// add the left child to
// the queue for further
// processing
if (temp.left != null)
{
q.add(temp.left);
}
// add the right child to
// the queue for further
// processing
if (temp.right != null)
{
q.add(temp.right);
}
}
}
// driver code
public static void main(string[] args)
{
string s = "geeksforgeeks";
node root = null;
// add individual characters
// to the string one by one
// in level order
for (int i = 0; i < s.length(); i )
{
root = addinlvlorder(root, s.charat(i));
}
// print the level order
// of the constructed
// binary tree
printlvlorder(root);
}
}
// this code is contributed by rajput-ji
c
// c# implementation of
// the above approach
using system;
using system.collections.generic;
class gfg
{
// node in the tree where
// data holds the character
// of the string and cnt
// holds the frequency
public class node
{
public char data;
public int cnt;
public node left, right;
};
// function to add a new
// node to the binary tree
static node add(char data)
{
// create a new node and
// populate its data part,
// set cnt as 1 and left
// and right children as null
node newnode = new node();
newnode.data = data;
newnode.cnt = 1;
newnode.left = newnode.right = null;
return newnode;
}
// function to add a node
// to the binary tree in
// level order
static node addinlvlorder(node root, char data)
{
if (root == null)
{
return add(data);
}
// use the queue data structure
// for level order insertion
// and push the root of tree to queue
list q = new list();
q.add(root);
while (q.count != 0)
{
node temp = q[0];
q.removeat(0);
// if the character to be
// inserted is present,
// update the cnt
if (temp.data == data)
{
temp.cnt ;
break;
}
// if the left child is
// empty add a new node
// as the left child
if (temp.left == null)
{
temp.left = add(data);
break;
}
else
{
// if the character is present
// as a left child, update the
// cnt and exit the loop
if (temp.left.data == data)
{
temp.left.cnt ;
break;
}
// add the left child to
// the queue for further
// processing
q.add(temp.left);
}
// if the right child is empty,
// add a new node to the right
if (temp.right == null)
{
temp.right = add(data);
break;
}
else
{
// if the character is present
// as a right child, update the
// cnt and exit the loop
if (temp.right.data == data)
{
temp.right.cnt ;
break;
}
// add the right child to
// the queue for further
// processing
q.add(temp.right);
}
}
return root;
}
// function to print the
// level order traversal of
// the binary tree
static void printlvlorder(node root)
{
// add the root to the queue
list q = new list();
q.add(root);
while (q.count != 0)
{
node temp = q[0];
// if the cnt of the character
// is more then one, display cnt
if (temp.cnt > 1)
{
console.write((temp.data "" temp.cnt));
}
// if the cnt of character
// is one, display character only
else
{
console.write((char)temp.data);
}
q.removeat(0);
// add the left child to
// the queue for further
// processing
if (temp.left != null)
{
q.add(temp.left);
}
// add the right child to
// the queue for further
// processing
if (temp.right != null)
{
q.add(temp.right);
}
}
}
// driver code
public static void main(string[] args)
{
string s = "geeksforgeeks";
node root = null;
// add individual characters
// to the string one by one
// in level order
for (int i = 0; i < s.length; i )
{
root = addinlvlorder(root, s[i]);
}
// print the level order
// of the constructed
// binary tree
printlvlorder(root);
}
}
// this code is contributed by rajput-ji
java 描述语言
output:
g2e4k2s2for
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