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

Difficulty Level : Basic
Last Updated : 30 Jun, 2021

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

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*/

Javascript

<script>
    // Javascript program to Check whether a given
    // binary tree is skewed binary tree or not?
     
    // Binary tree node
    class Node
    {
        constructor(key) {
           this.left = null;
           this.right = null;
           this.data = key;
        }
    }
     
    // Utility function to create a new node
    function newNode(key)
    {
        let temp = new Node(key);
        return (temp);
    }
 
    function isSkewedBT(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);
    }
     
    /* 10
    /     \
    12     13
        /     \
        14     15    
        / \     / \
    21 22 23 24        
    Let us create Binary Tree shown in above example */
    let root = newNode(10);
    root.left = newNode(12);
    root.left.right = newNode(15);
    document.write(isSkewedBT(root)?1 + "</br>":0 + "</br>");
   
    root = newNode(5);
    root.right = newNode(4);
    root.right.left = newNode(3);
    root.right.left.right = newNode(2);
    document.write(isSkewedBT(root)?1 + "</br>":0 + "</br>");
   
    root = newNode(5);
    root.left = newNode(4);
    root.left.right = newNode(3);
    root.left.right.left = newNode(2);
    root.left.right.right = newNode(4);
    document.write(isSkewedBT(root)?1 + "</br>":0 + "</br>");
  
 // This code is contributed by decode2207.
</script>

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes