Skip to content
Related Articles

Related Articles

Improve Article

Check if a given Binary Tree is height balanced like a Red-Black Tree

  • Difficulty Level : Medium
  • Last Updated : 02 Jun, 2021

In a Red-Black Tree, the maximum height of a node is at most twice the minimum height (The four Red-Black tree properties make sure this is always followed). Given a Binary Search Tree, we need to check for following property. 
For every node, length of the longest leaf to node path has not more than twice the nodes on shortest path from node to leaf.
 

    12                                        40
      \                                     /    \ 
       14                                 10      100    
         \                                        /  \
          16                                     60   150    
 Cannot be a Red-Black Tree              It can be Red-Black Tree
 with any color assignment
 Max height of 12 is 1
 Min height of 12 is 3


          10
        /   \
      5     100
           /   \
          50   150
         /
        40 
 It can also be Red-Black Tree

Expected time complexity is O(n). The tree should be traversed at-most once in the solution.
We strongly recommend to minimize the browser and try this yourself first. 
For every node, we need to get the maximum and minimum heights and compare them. The idea is to traverse the tree and for every node check if it’s balanced. We need to write a recursive function that returns three things, a boolean value to indicate the tree is balanced or not, minimum height and maximum height. To return multiple values, we can either use a structure or pass variables by reference. We have passed maxh and minh by reference so that the values can be used in parent calls. 
 

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.

C++




/* Program to check if a given Binary Tree is balanced like a Red-Black Tree */
#include <bits/stdc++.h>
using namespace std;
 
struct Node
{
    int key;
    Node *left, *right;
};
 
/* utility that allocates a new Node with the given key  */
Node* newNode(int key)
{
    Node* node = new Node;
    node->key = key;
    node->left = node->right = NULL;
    return (node);
}
 
// Returns returns tree if the Binary tree is balanced like a Red-Black
// tree. This function also sets value in maxh and minh (passed by
// reference). maxh and minh are set as maximum and minimum heights of root.
bool isBalancedUtil(Node *root, int &maxh, int &minh)
{
    // Base case
    if (root == NULL)
    {
        maxh = minh = 0;
        return true;
    }
 
    int lmxh, lmnh; // To store max and min heights of left subtree
    int rmxh, rmnh; // To store max and min heights of right subtree
 
    // Check if left subtree is balanced, also set lmxh and lmnh
    if (isBalancedUtil(root->left, lmxh, lmnh) == false)
        return false;
 
    // Check if right subtree is balanced, also set rmxh and rmnh
    if (isBalancedUtil(root->right, rmxh, rmnh) == false)
        return false;
 
    // Set the max and min heights of this node for the parent call
    maxh = max(lmxh, rmxh) + 1;
    minh = min(lmnh, rmnh) + 1;
 
    // See if this node is balanced
    if (maxh <= 2*minh)
        return true;
 
    return false;
}
 
// A wrapper over isBalancedUtil()
bool isBalanced(Node *root)
{
    int maxh, minh;
    return isBalancedUtil(root, maxh, minh);
}
 
/* Driver program to test above functions*/
int main()
{
    Node * root = newNode(10);
    root->left = newNode(5);
    root->right = newNode(100);
    root->right->left = newNode(50);
    root->right->right = newNode(150);
    root->right->left->left = newNode(40);
    isBalanced(root)? cout << "Balanced" : cout << "Not Balanced";
 
    return 0;
}

Java




// Java Program to check if a given Binary
// Tree is balanced like a Red-Black Tree
class GFG
{
 
static class Node
{
    int key;
    Node left, right;
    Node(int key)
    {
        left = null;
        right = null;
        this.key = key;
    }
}
 
static class INT
{
    static int d;
    INT()
    {
        d = 0;
    }
}
 
// Returns returns tree if the Binary
// tree is balanced like a Red-Black
// tree. This function also sets value
// in maxh and minh (passed by reference).
// maxh and minh are set as maximum and
// minimum heights of root.
static boolean isBalancedUtil(Node root,
                        INT maxh, INT minh)
{
     
    // Base case
    if (root == null)
    {
        maxh.d = minh.d = 0;
        return true;
    }
     
    // To store max and min heights of left subtree
    INT lmxh=new INT(), lmnh=new INT();
     
    // To store max and min heights of right subtree
    INT rmxh=new INT(), rmnh=new INT();
 
    // Check if left subtree is balanced,
    // also set lmxh and lmnh
    if (isBalancedUtil(root.left, lmxh, lmnh) == false)
        return false;
 
    // Check if right subtree is balanced,
    // also set rmxh and rmnh
    if (isBalancedUtil(root.right, rmxh, rmnh) == false)
        return false;
 
    // Set the max and min heights
    // of this node for the parent call
    maxh.d = Math.max(lmxh.d, rmxh.d) + 1;
    minh.d = Math.min(lmnh.d, rmnh.d) + 1;
 
    // See if this node is balanced
    if (maxh.d <= 2*minh.d)
        return true;
 
    return false;
}
 
// A wrapper over isBalancedUtil()
static boolean isBalanced(Node root)
{
    INT maxh=new INT(), minh=new INT();
    return isBalancedUtil(root, maxh, minh);
}
 
// Driver code
public static void main(String args[])
{
    Node root = new Node(10);
    root.left = new Node(5);
    root.right = new Node(100);
    root.right.left = new Node(50);
    root.right.right = new Node(150);
    root.right.left.left = new Node(40);
    System.out.println(isBalanced(root) ?
            "Balanced" : "Not Balanced");
}
}
 
// This code is contributed by Arnab Kundu

Python3




""" Program to check if a given Binary
Tree is balanced like a Red-Black Tree """
 
# Helper function that allocates a new
# node with the given data and None
# left and right poers.                                
class newNode:
 
    # Construct to create a new node
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# Returns returns tree if the Binary
# tree is balanced like a Red-Black
# tree. This function also sets value
# in maxh and minh (passed by
# reference). maxh and minh are set
# as maximum and minimum heights of root.
def isBalancedUtil(root, maxh, minh) :
 
    # Base case
    if (root == None) :
     
        maxh = minh = 0
        return True
     
 
    lmxh=0
     
    # To store max and min
    # heights of left subtree
    lmnh=0
     
    # To store max and min
    # heights of right subtree
    rmxh, rmnh=0,0 
 
    # Check if left subtree is balanced,
    # also set lmxh and lmnh
    if (isBalancedUtil(root.left, lmxh, lmnh) == False) :
        return False
 
    # Check if right subtree is balanced,
    # also set rmxh and rmnh
    if (isBalancedUtil(root.right, rmxh, rmnh) == False) :
        return False
 
    # Set the max and min heights of
    # this node for the parent call
    maxh = max(lmxh, rmxh) + 1
    minh = min(lmnh, rmnh) + 1
 
    # See if this node is balanced
    if (maxh <= 2 * minh) :
        return True
 
    return False
 
 
# A wrapper over isBalancedUtil()
def isBalanced(root) :
 
    maxh, minh =0,0
    return isBalancedUtil(root, maxh, minh)
 
# Driver Code
if __name__ == '__main__':
    root = newNode(10)
    root.left = newNode(5)
    root.right = newNode(100)
    root.right.left = newNode(50)
    root.right.right = newNode(150)
    root.right.left.left = newNode(40)
    if (isBalanced(root)):
        print("Balanced")
    else:
        print("Not Balanced")
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# Program to check if a given Binary
// Tree is balanced like a Red-Black Tree
using System;
 
class GFG
{
 
public class Node
{
    public int key;
    public Node left, right;
    public Node(int key)
    {
        left = null;
        right = null;
        this.key = key;
    }
}
 
public class INT
{
    public int d;
    public INT()
    {
        d = 0;
    }
}
 
// Returns returns tree if the Binary
// tree is balanced like a Red-Black
// tree. This function also sets value
// in maxh and minh (passed by reference).
// maxh and minh are set as maximum and
// minimum heights of root.
static bool isBalancedUtil(Node root,
                        INT maxh, INT minh)
{
     
    // Base case
    if (root == null)
    {
        maxh.d = minh.d = 0;
        return true;
    }
     
    // To store max and min heights of left subtree
    INT lmxh = new INT(), lmnh = new INT();
     
    // To store max and min heights of right subtree
    INT rmxh = new INT(), rmnh = new INT();
 
    // Check if left subtree is balanced,
    // also set lmxh and lmnh
    if (isBalancedUtil(root.left, lmxh, lmnh) == false)
        return false;
 
    // Check if right subtree is balanced,
    // also set rmxh and rmnh
    if (isBalancedUtil(root.right, rmxh, rmnh) == false)
        return false;
 
    // Set the max and min heights
    // of this node for the parent call
    maxh.d = Math.Max(lmxh.d, rmxh.d) + 1;
    minh.d = Math.Min(lmnh.d, rmnh.d) + 1;
 
    // See if this node is balanced
    if (maxh.d <= 2 * minh.d)
        return true;
 
    return false;
}
 
// A wrapper over isBalancedUtil()
static bool isBalanced(Node root)
{
    INT maxh = new INT(), minh = new INT();
    return isBalancedUtil(root, maxh, minh);
}
 
// Driver code
public static void Main(String []args)
{
    Node root = new Node(10);
    root.left = new Node(5);
    root.right = new Node(100);
    root.right.left = new Node(50);
    root.right.right = new Node(150);
    root.right.left.left = new Node(40);
    Console.WriteLine(isBalanced(root) ?
            "Balanced" : "Not Balanced");
}
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript Program to check if a given Binary
// Tree is balanced like a Red-Black Tree    
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
     class INT {
         constructor() {
            this.d = 0;
        }
    }
 
    // Returns returns tree if the Binary
    // tree is balanced like a Red-Black
    // tree. This function also sets value
    // in maxh and minh (passed by reference).
    // maxh and minh are set as maximum and
    // minimum heights of root.
    function isBalancedUtil(root,  maxh,  minh) {
 
        // Base case
        if (root == null) {
            maxh.d = minh.d = 0;
            return true;
        }
 
        // To store max and min heights of left subtree
        var lmxh = new INT(), lmnh = new INT();
 
        // To store max and min heights of right subtree
        var rmxh = new INT(), rmnh = new INT();
 
        // Check if left subtree is balanced,
        // also set lmxh and lmnh
        if (isBalancedUtil(root.left, lmxh, lmnh) == false)
            return false;
 
        // Check if right subtree is balanced,
        // also set rmxh and rmnh
        if (isBalancedUtil(root.right, rmxh, rmnh) == false)
            return false;
 
        // Set the max and min heights
        // of this node for the parent call
        maxh.d = Math.max(lmxh.d, rmxh.d) + 1;
        minh.d = Math.min(lmnh.d, rmnh.d) + 1;
 
        // See if this node is balanced
        if (maxh.d <= 2 * minh.d)
            return true;
 
        return false;
    }
 
    // A wrapper over isBalancedUtil()
    function isBalanced(root) {
        var maxh = new INT(), minh = new INT();
        return isBalancedUtil(root, maxh, minh);
    }
 
    // Driver code
     
        var root = new Node(10);
        root.left = new Node(5);
        root.right = new Node(100);
        root.right.left = new Node(50);
        root.right.right = new Node(150);
        root.right.left.left = new Node(40);
        document.write(isBalanced(root) ?
        "Balanced" : "Not Balanced");
         
// This code contributed by umadevi9616
 
</script>

Output: 

Balanced 

Time Complexity: Time Complexity of above code is O(n) as the code does a simple tree traversal.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :