You are given the root of a complete binary tree, you need to calculate the number of nodes in the given complete binary tree.
A complete binary tree is a tree in which all the levels are completely filled except the last level. Nodes in the last level are as left as possible.
For Example:
In the above complete binary tree, all the levels are filled except for the last. In the last level, all the nodes in the last level are as far left as possible.
The first line contains an Integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the above image would be:
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation:
Level 1 :
The root node of the tree is 1.
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Return the number of nodes of the given complete
binary tree.
Constraints:
1 <= T <= 10
0 <= Number of nodes in tree <= 10^5
1 <= Nodes Value <= 5*10^5
Time Limit: 1 second
1
1 2 3 4 5 7 6 -1 -1 -1 -1 -1 -1 -1 -1
7
For the above test case, the given tree is:
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = 5
Left child of 3 = 7
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = null(-1)
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 7 = null (-1)
Right child of 7 = null (-1).
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Hence the total number of nodes in the given tree is 7.
1
5 6 7 10 -1 -1 -1 -1 -1
4
Traverse all the nodes of the tree.
Explanation:
The key idea is to traverse all the nodes of the tree and return the count of the number of nodes traversed.
Algorithm:
O(N), Where ‘N’ is the number of nodes in the binary tree.
We traverse all the nodes. Hence time complexity will be O(N).
O(log(N)), Where ‘N’ is the number of nodes in the binary tree.
The space required for the recursion call stack will be equal to the height of the complete binary tree which is equal to log(N).
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Interview problems
can anyone tell me where is failed my code
#include <bits/stdc++.h>
/*************************************************************
Following is the Binary Tree node structure
class BinaryTreeNode {
public :
T data;
BinaryTreeNode<T> *left;
BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this -> data = data;
left = NULL;
right = NULL;
}
};
*************************************************************/
int countNodes(BinaryTreeNode<int> *root) {
if(!root)return 0;
return 1+countNodes(root->left)+countNodes(root->right);
}Interview problems
Java straightforward solution
import java.util.* ;
import java.io.*;
/*************************************************************
Following is the Binary Tree node structure
class BinaryTreeNode<T> {
public T data;
public BinaryTreeNode<T> *left;
public BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this.data = data;
left = NULL;
right = NULL;
}
};
*************************************************************/
public class Solution {
public static int countNodes(BinaryTreeNode<Integer> root) {
// Write your code here.
Queue<BinaryTreeNode<Integer>> q = new LinkedList<>();
q.add(root);
int i = 1;
while(!q.isEmpty()){
BinaryTreeNode node = q.poll();
if(node.left == null && node.right == null){
continue;
}
if(node.left != null){
q.add(node.left);
i++;
}
if(node.right != null){
q.add(node.right);
i++;
}
}
return i;
}
}Interview problems
C++ || Easy solution
#include <bits/stdc++.h>
int height(BinaryTreeNode<int> *root, bool l){
int h = 0;
while(root){
h++;
if(l) root = root->left;
else root = root->right;
}
return h;
}
int countNodes(BinaryTreeNode<int> *root) {
int leftH = height(root, true);
int rightH = height(root, false);
if(leftH == rightH) return ((1 << leftH) - 1);
return 1 + countNodes(root->left) + countNodes(root->right);
}
Interview problems
TLE
can someone let me know why does this cause TLE??
#include <bits/stdc++.h>
/*************************************************************
Following is the Binary Tree node structure
class BinaryTreeNode {
public :
T data;
BinaryTreeNode<T> *left;
BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this -> data = data;
left = NULL;
right = NULL;
}
};
*************************************************************/
int countNodes(BinaryTreeNode<int> *root) {
queue<BinaryTreeNode<int>*>q;
if(root==NULL) return 0;
q.push(root);
int ans=0;
while(!q.empty()){
auto it=q.front();
ans++;
q.pop();
if(it->left) q.push(it->left);
if(it->right) q.push(it->right);
}
return ans;
}Interview problems
C++ Solution
int findLeftHeight(BinaryTreeNode<int>* root)
{
int h=0;
while(root)
{
h++;
root=root->left;
}
return h;
}
int findRightHeight(BinaryTreeNode<int>* root)
{
int h=0;
while(root)
{
h++;
root=root->right;
}
return h;
}
int countNodes(BinaryTreeNode<int> *root) {
// Write your code here.
if(!root)
return 0;
int lh=findLeftHeight(root);
int rh=findRightHeight(root);
if(lh==rh)
return ((1<<lh)-1);
return 1+countNodes(root->left)+countNodes(root->right);
}
Interview problems
EASY JAVA SOLUTION
import java.util.* ;
public class Solution { public static int nodes(BinaryTreeNode<Integer> root){ if(root == null) return 0; int left=nodes(root.left); int right=nodes(root.right); return 1+right+left; } public static int countNodes(BinaryTreeNode<Integer> root) { int cnt=nodes(root); return cnt; } }
Interview problems
best solution in c++
#include <bits/stdc++.h>
/*************************************************************
Following is the Binary Tree node structure
class BinaryTreeNode {
public :
T data;
BinaryTreeNode<T> *left;
BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this -> data = data;
left = NULL;
right = NULL;
}
};
*************************************************************/
int findlefthight(BinaryTreeNode<int>* node){
int hight = 0;
while(node){
hight++;
node = node->left;
}
return hight;
}
int findrightheight(BinaryTreeNode<int>* node){
int hight = 0;
while(node){
hight++;
node = node->right;
}
return hight;
}
int countNodes(BinaryTreeNode<int> *root) {
// Write your code here.
if(root == NULL)
return 0;
int lh = findlefthight(root);
int rh = findrightheight(root);
if(lh == rh) return (1<<lh)-1 ;
int leftNodes = countNodes(root->left);
int rightNodes = countNodes(root->right);
return 1 + leftNodes + rightNodes;
}
Interview problems
c++ easiest
int height(BinaryTreeNode<int>*r,bool l)
{
int c=0;
while(r)
{
c++;
if(l) r=r->left;
else r=r->right;
}
return c;
}
int countNodes(BinaryTreeNode<int>*r) {
if(r==NULL)return 0;
int lh=height(r,true);
int rh=height(r,false);
if(lh==rh)return pow(2,lh)-1;
return 1+countNodes(r->left)+countNodes(r->right);
}Interview problems
Easy C++ Code With log(N) time complexity
#include <bits/stdc++.h>
/*************************************************************
Following is the Binary Tree node structure
class BinaryTreeNode {
public :
T data;
BinaryTreeNode<T> *left;
BinaryTreeNode<T> *right;
BinaryTreeNode(T data) {
this -> data = data;
left = NULL;
right = NULL;
}
};
*************************************************************/
int leftheight(BinaryTreeNode<int>* node){
int h=0;
while(node){
h++;
node=node->left;
}
return h;
}
int rightheight(BinaryTreeNode<int>* node){
int h=0;
while(node){
h++;
node=node->right;
}
return h;
}
int countNodes(BinaryTreeNode<int> *root) {
// Write your code here.
if(root == NULL){
return 0;
}
int lh= leftheight(root);
int rh= rightheight(root);
if(lh == rh){
return (pow(2,lh)-1);
}
return 1+countNodes(root->left)+countNodes(root->right);
}
Interview problems
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