Check whether a given binary tree is skewed binary tree or not?

Given a Binary Tree check whether it is skewed binary tree or not. A skewed tree is a tree where each node has only one child node or none.

Examples:

Input :         5
             /   
            4
              \
               3
             /
           2
Output : Yes

Input :       5
            /
           4
            \
             3
           /  \
          2    4
Output : No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to check if a node has two children. If node has two children return false, else recursively compute whether subtree with one child is skewed tree. If node is leaf node return true.

Below is the implementation of above approach:

C++

// C++ program to Check whether a given  
// binary tree is skewed binary tree or not? 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// A Tree node 
struct Node { 
    int key; 
    struct Node *left, *right; 
}; 
  
// Utility function to create a new node 
Node* newNode(int key) 
{ 
    Node* temp = new Node; 
    temp->key = key; 
    temp->left = temp->right = NULL; 
    return (temp); 
} 
  
bool isSkewedBT(Node* root) 
{ 
    // check if node is NULL or is a leaf node 
    if (root == NULL || (root->left == NULL &&  
                        root->right == NULL)) 
        return true; 
  
    // check if node has two children if 
    // yes, return false 
    if (root->left && root->right)  
        return false; 
    if (root->left) 
        return isSkewedBT(root->left); 
    return isSkewedBT(root->right); 
} 
  
// Driver program 
int main() 
{ 
    /*   10 
       /     \ 
     12       13 
           /     \ 
         14       15     
        /   \     /  \ 
       21   22   23   24            
    Let us create Binary Tree shown in above example */
    Node* root = newNode(10); 
    root->left = newNode(12); 
    root->left->right = newNode(15); 
    cout << isSkewedBT(root) << endl; 
  
    root = newNode(5); 
    root->right = newNode(4); 
    root->right->left = newNode(3); 
    root->right->left->right = newNode(2); 
    cout << isSkewedBT(root) << endl; 
  
    root = newNode(5); 
    root->left = newNode(4); 
    root->left->right = newNode(3); 
    root->left->right->left = newNode(2); 
    root->left->right->right = newNode(4); 
    cout << isSkewedBT(root) << endl; 
} 

Java

// Java program to Check whether a given  
// binary tree is skewed binary tree or not? 
  
class Solution 
{ 
// A Tree node  
 static class Node {  
    int key;  
     Node left, right;  
} 
  
// Utility function to create a new node  
static Node newNode(int key)  
{  
    Node temp = new Node();  
    temp.key = key;  
    temp.left = temp.right = null;  
    return (temp);  
}  
  
static boolean isSkewedBT(Node root)  
{  
    // check if node is null or is a leaf node  
    if (root == null || (root.left == null &&  
                        root.right == null))  
        return true;  
  
    // check if node has two children if  
    // yes, return false  
    if (root.left!=null && root.right!=null)  
        return false;  
    if (root.left!=null)  
        return isSkewedBT(root.left);  
    return isSkewedBT(root.right);  
}  
  
// Driver program  
public static void main(String args[]) 
{  
    /* 10  
    /     \  
    12     13  
        /     \  
        14     15      
        / \     / \  
    21 22 23 24          
    Let us create Binary Tree shown in above example */
    Node root = newNode(10);  
    root.left = newNode(12);  
    root.left.right = newNode(15);  
    System.out.println( isSkewedBT(root)?1:0 );  
  
    root = newNode(5);  
    root.right = newNode(4);  
    root.right.left = newNode(3);  
    root.right.left.right = newNode(2);  
    System.out.println( isSkewedBT(root)?1:0 );  
  
    root = newNode(5);  
    root.left = newNode(4);  
    root.left.right = newNode(3);  
    root.left.right.left = newNode(2);  
    root.left.right.right = newNode(4);  
    System.out.println(  isSkewedBT(root)?1:0 );  
}  
} 
//contributed by Arnab Kundu 

Python3

# Python3 program to Check whether a given  
# binary tree is skewed binary tree or not?  
  
# Binary Tree Node  
""" utility that allocates a newNode  
with the given key """
class newNode:  
  
    # Construct to create a newNode  
    def __init__(self, key):  
        self.data = key 
        self.left = None
        self.right = None
  
def isSkewedBT(root):  
  
    # check if node is None or is a leaf node  
    if (root == None or (root.left == None and
                         root.right == None)):  
        return 1
  
    # check if node has two children if  
    # yes, return false  
    if (root.left and root.right):  
        return 0
    if (root.left) : 
        return isSkewedBT(root.left)  
    return isSkewedBT(root.right)  
  
# Driver Code  
if __name__ == '__main__': 
  
    """ 10  
    /     \  
    12     13  
        /     \  
        14     15      
        / \     / \  
    21 22 23 24          
    Let us create Binary Tree shown in above example """
    root = newNode(10)  
    root.left = newNode(12)  
    root.left.right = newNode(15)  
    print(isSkewedBT(root)) 
  
    root = newNode(5)  
    root.right = newNode(4)  
    root.right.left = newNode(3)  
    root.right.left.right = newNode(2)  
    print(isSkewedBT(root)) 
  
    root = newNode(5)  
    root.left = newNode(4)  
    root.left.right = newNode(3)  
    root.left.right.left = newNode(2)  
    root.left.right.right = newNode(4)  
    print(isSkewedBT(root)) 
  
# This code is contributed by 
# Shubham Singh(SHUBHAMSINGH10) 

C#

// C# program to Check whether a given  
// binary tree is skewed binary tree or not?  
using System; 
  
public class GFG  
{  
      
// A Tree node  
class Node  
{  
    public int key;  
    public Node left, right;  
}  
  
// Utility function to create a new node  
static Node newNode(int key)  
{  
    Node temp = new Node();  
    temp.key = key;  
    temp.left = temp.right = null;  
    return (temp);  
}  
  
static bool isSkewedBT(Node root)  
{  
    // check if node is null or is a leaf node  
    if (root == null || (root.left == null &&  
                        root.right == null))  
        return true;  
  
    // check if node has two children if  
    // yes, return false  
    if (root.left!=null && root.right!=null)  
        return false;  
    if (root.left!=null)  
        return isSkewedBT(root.left);  
    return isSkewedBT(root.right);  
}  
  
// Driver code  
public static void Main()  
{  
    /* 10  
    / \  
    12 13  
        / \  
        14 15  
        / \ / \  
    21 22 23 24  
    Let us create Binary Tree shown in above example */
    Node root = newNode(10);  
    root.left = newNode(12);  
    root.left.right = newNode(15);  
    Console.WriteLine( isSkewedBT(root)?1:0 );  
  
    root = newNode(5);  
    root.right = newNode(4);  
    root.right.left = newNode(3);  
    root.right.left.right = newNode(2);  
    Console.WriteLine( isSkewedBT(root)?1:0 );  
  
    root = newNode(5);  
    root.left = newNode(4);  
    root.left.right = newNode(3);  
    root.left.right.left = newNode(2);  
    root.left.right.right = newNode(4);  
    Console.WriteLine( isSkewedBT(root)?1:0 );  
}  
}  
  
/* This code is contributed by Rajput-Ji*/

Output:

1
1
0

Time complexity of this solution is
Best case : O(1) when root has two children.
Worst case : O(h) when tree is skewed tree.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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