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Binary tree

Anxious Armadillo answered on March 16, 2021 Popularity 10/10 Helpfulness 1/10

answer Binary tree

related binary tree

Binary tree

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void preorder(node *n) {
    cout << n->value;
    if(n->left!=NULL)
        preorder(n->left);
    if (n->right != NULL)
        preorder(n->right);
}
void inorder(node *n) {
    if (n->left != NULL)
        inorder(n->left);
    cout << n->value;
    if (n->right != NULL)
        inorder(n->right);
}
void postorder(node *n) {
    if (n->left != NULL)
        postorder(n->left);
    if (n->right != NULL)
        postorder(n->right);
    cout << n->value;
}

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: minimin2.tistory.com

Tags: binary-tree whatever

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Contributed on May 27 2022

Anxious Armadillo

0 Answers Avg Quality 2/10

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binary tree

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class Node:
    def __init__(self,key):
        self.left = None
        self.right = None
        self.val = key
 
root = Node(1)
root.left      = Node(2);
root.right     = Node(3);
root.left.left  = Node(4);

Popularity 10/10 Helpfulness 2/10 Language python

Source: Grepper

Tags: binary-tree bina

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Contributed on Mar 16 2021

Important Impala

0 Answers Avg Quality 2/10

binary tree

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#include <iostream>
#include <queue>
using namespace std;
struct Node{
   int data;
   struct Node* left, *right;
};
// Function to count the full Nodes in a binary tree
int fullcount(struct Node* node){
   // Check if tree is empty
   if (!node){
      return 0;
   }  
   queue<Node *> myqueue;
   // traverse using level order traversing
   int result = 0;
   myqueue.push(node);
   while (!myqueue.empty()){
      struct Node *temp = myqueue.front();
      myqueue.pop();
      if (temp->left && temp->right){
         result++;
      }
      if (temp->left != NULL){
         myqueue.push(temp->left);
      }
      if (temp->right != NULL){
         myqueue.push(temp->right);
      }
   }
   return result;
}
struct Node* newNode(int data){
   struct Node* node = new Node;
   node->data = data;
   node->left = node->right = NULL;
   return (node);
}
int main(void){
   struct Node *root = newNode(10);
   root->left = newNode(20);
   root->right = newNode(30);
   root->left->left = newNode(40);
   root->left->right = newNode(50);
   root->left->left->right = newNode(60);
   root->left->right->right = newNode(70);
   cout <<"count is: "<<fullcount(root);
   return 0;
}

Popularity 10/10 Helpfulness 1/10 Language cpp

Source: Grepper

Tags: binary-tree bina

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Contributed on Feb 16 2022

Mohith

0 Answers Avg Quality 2/10

binary tree

Comment

void preorder(node *n) {
    cout << n->value;
    if(n->left!=NULL)
        preorder(n->left);
    if (n->right != NULL)
        preorder(n->right);
}
void inorder(node *n) {
    if (n->left != NULL)
        inorder(n->left);
    cout << n->value;
    if (n->right != NULL)
        inorder(n->right);
}
void postorder(node *n) {
    if (n->left != NULL)
        postorder(n->left);
    if (n->right != NULL)
        postorder(n->right);
    cout << n->value;
}

Popularity 10/10 Helpfulness 1/10 Language cpp

Source: Grepper

Tags: binary-tree bina

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Contributed on May 27 2022

Anxious Armadillo

0 Answers Avg Quality 2/10

binary tree

Comment

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: leetcode.com

Tags: binary-tree bina

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Contributed on Jun 24 2022

shyam makwana

0 Answers Avg Quality 2/10

binary tree

Comment

#include <bits/stdc++.h>
using namespace std;
  
class Node {
  public:
    int data;
    Node* left;
    Node* right;
  
    // Val is the key or the value that
    // has to be added to the data part
    Node(int val)
    {
        data = val;
  
        // Left and right child for node
        // will be initialized to null
        left = NULL;
        right = NULL;
    }
};
  
int main()
{
  
    /*create root*/
    Node* root = new Node(1);
    /* following is the tree after above statement
  
             1
            / \
        NULL   NULL
    */
  
    root->left = new Node(2);
    root->right = new Node(3);
    /* 2 and 3 become left and right children of 1
                  1
                /   \
                2      3
               / \     / \
            NULL NULL NULL NULL
    */
  
    root->left->left = new Node(4);
    /* 4 becomes left child of 2
              1
            /    \
           2      3
          / \     / \
         4 NULL NULL NULL
        / \
      NULL NULL
    */
  
    return 0;
}

Popularity 9/10 Helpfulness 1/10 Language cpp

Source: www.geeksforgeeks.org

Tags: binary-tree bina

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Contributed on Sep 18 2022

Arnold Moyo

0 Answers Avg Quality 2/10

Binary tree

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