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.

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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.

/* Program to check if a given Binary Tree is balanced like a Red-Black Tree */ #include <iostream> 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; }

Output:

Balanced

Time Complexity: Time Complexity of above code is O(n) as the code does a simple tree traversal.

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