Count levels in a Binary Tree consisting of node values having set bits at different positions
Given a Binary Tree consisting of N nodes, the task is to count the number of levels in a Binary Tree such that the set bits of all the node values at the same level is at different positions.
Examples:
Input:
5 / \ 6 9 / \ \ 1 4 7Output: 2
Explanation:
Level 1 has only 5 (= (101)2).
Level 2 has 6 (= (0110)2) and 9 (= (1001)2). All set bits are at unique positions.
Level 3 has 1 (0001)2, 4 (0100)2 and 7(0111)2. Therefore, 0th bit of node values 5 and 7 are set.Input:
1 / \ 2 3 / \ \ 5 4 7Output: 1
Naive Approach: The simplest approach to solve this problem to traverse the binary tree using level order traversal and at each level of the tree store the set bits of all the nodes using Map. Traverse the map and check if the frequency of set-bit at the same position is less than or equal to 1 or not. If found to be true, then increment the count. Finally, print the count obtained.
Time Complexity: O(N)
Auxiliary Space: O(32)
Efficient Approach: The above approach can be optimized based on the following observations:
If all the set bits of two numbers A and B are at different positions
A XOR B = A OR B
Follow the steps below to solve the problem:
- Initialize a variable, say prefiX_XOR, to store the prefix XOR of all the nodes at each level.
- Initialize a variable, say prefiX_OR, to store the prefix OR of all the nodes at each level.
- Traverse the binary tree using level order traversal. At every ith level, check if prefix_XOR ^ nodes is equal to (prefix_OR | nodes) or not. If found to be true for all the nodes at current level, then increment the count.
- Finally, print the count obtained.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Structure of a node in // the binary tree struct TreeNode { int val = 0; TreeNode *left,*right; TreeNode( int x) { val = x; left = NULL; right = NULL; } }; // Function to find total unique levels void uniqueLevels(TreeNode *root) { // Stores count of levels, where the set // bits of all the nodes are at // different positions int uniqueLevels = 0; // Store nodes at each level of // the tree using BFS queue<TreeNode*> que; que.push(root); // Performing level order traversal while (que.size() > 0) { // Stores count of nodes at // current level int length = que.size(); // Stores prefix XOR of all // the nodes at current level int prefix_XOR = 0; // Stores prefix OR of all // the nodes at current level int prefix_OR = 0; // Check if set bit of all the nodes // at current level is at different // positions or not bool flag = true ; // Traverse nodes at current level for ( int i = 0; i < length; i++){ // Stores front element // of the que TreeNode *temp = que.front(); que.pop(); // Update prefix_OR prefix_OR |= temp->val; // Update prefix_XOR prefix_XOR ^= temp->val; if (prefix_XOR != prefix_OR) flag = false ; // If left subtree not NULL if (temp->left) que.push(temp->left); // If right subtree not NULL if (temp->right) que.push(temp->right); // Update length } //If bitwise AND is zero if (flag) uniqueLevels += 1; } cout << uniqueLevels; } // Driver Code int main() { TreeNode *root = new TreeNode(5); root->left = new TreeNode(6); root->right = new TreeNode(9); root->left->left = new TreeNode(1); root->left->right = new TreeNode(4); root->right->right = new TreeNode(7); // Function Call uniqueLevels(root); return 0; } // This code is contributed by mohit kumar 29. |
Java
// Java program for the above approach import java.util.*; class GFG { // Structure of a node in // the binary tree static class TreeNode { int val = 0 ; TreeNode left, right; TreeNode( int x) { val = x; left = null ; right = null ; } }; // Function to find total unique levels static void uniqueLevels(TreeNode root) { // Stores count of levels, where the set // bits of all the nodes are at // different positions int uniqueLevels = 0 ; // Store nodes at each level of // the tree using BFS Queue<TreeNode> que = new LinkedList<>(); que.add(root); // Performing level order traversal while (que.size() > 0 ) { // Stores count of nodes at // current level int length = que.size(); // Stores prefix XOR of all // the nodes at current level int prefix_XOR = 0 ; // Stores prefix OR of all // the nodes at current level int prefix_OR = 0 ; // Check if set bit of all the nodes // at current level is at different // positions or not boolean flag = true ; // Traverse nodes at current level for ( int i = 0 ; i < length; i++) { // Stores front element // of the que TreeNode temp = que.peek(); que.remove(); // Update prefix_OR prefix_OR |= temp.val; // Update prefix_XOR prefix_XOR ^= temp.val; if (prefix_XOR != prefix_OR) flag = false ; // If left subtree not null if (temp.left != null ) que.add(temp.left); // If right subtree not null if (temp.right != null ) que.add(temp.right); // Update length } //If bitwise AND is zero if (flag) uniqueLevels += 1 ; } System.out.print(uniqueLevels); } // Driver Code public static void main(String[] args) { TreeNode root = new TreeNode( 5 ); root.left = new TreeNode( 6 ); root.right = new TreeNode( 9 ); root.left.left = new TreeNode( 1 ); root.left.right = new TreeNode( 4 ); root.right.right = new TreeNode( 7 ); // Function Call uniqueLevels(root); } } // This code is contributed by 29AjayKumar |
Python3
# Python program for the above approach # Structure of a node in # the binary tree class TreeNode: def __init__( self , val = 0 , left = None , right = None ): self .val = val self .left = left self .right = right # Function to find total unique levels def uniqueLevels(root): # Stores count of levels, where the set # bits of all the nodes are at # different positions uniqueLevels = 0 # Store nodes at each level of # the tree using BFS que = [root] # Performing level order traversal while len (que): # Stores count of nodes at # current level length = len (que) # Stores prefix XOR of all # the nodes at current level prefix_XOR = 0 ; # Stores prefix OR of all # the nodes at current level prefix_OR = 0 # Check if set bit of all the nodes # at current level is at different # positions or not flag = True # Traverse nodes at current level while length: # Stores front element # of the que temp = que.pop( 0 ) # Update prefix_OR prefix_OR | = temp.val # Update prefix_XOR prefix_XOR ^ = temp.val if prefix_XOR ! = prefix_OR: flag = False # If left subtree not NULL if temp.left: que.append(temp.left) # If right subtree not NULL if temp.right: que.append(temp.right) # Update length length - = 1 # If bitwise AND is zero if flag: uniqueLevels + = 1 print (uniqueLevels) # Driver Code if __name__ = = '__main__' : root = TreeNode( 5 ) root.left = TreeNode( 6 ) root.right = TreeNode( 9 ) root.left.left = TreeNode( 1 ) root.left.right = TreeNode( 4 ) root.right.right = TreeNode( 7 ) # Function Call uniqueLevels(root) |
C#
// C# program for the above approach using System; using System.Collections.Generic; public class GFG { // Structure of a node in // the binary tree class TreeNode { public int val = 0; public TreeNode left, right; public TreeNode( int x) { val = x; left = null ; right = null ; } }; // Function to find total unique levels static void uniqueLevels(TreeNode root) { // Stores count of levels, where the set // bits of all the nodes are at // different positions int uniqueLevels = 0; // Store nodes at each level of // the tree using BFS Queue<TreeNode> que = new Queue<TreeNode>(); que.Enqueue(root); // Performing level order traversal while (que.Count > 0) { // Stores count of nodes at // current level int length = que.Count; // Stores prefix XOR of all // the nodes at current level int prefix_XOR = 0; // Stores prefix OR of all // the nodes at current level int prefix_OR = 0; // Check if set bit of all the nodes // at current level is at different // positions or not bool flag = true ; // Traverse nodes at current level for ( int i = 0; i < length; i++) { // Stores front element // of the que TreeNode temp = que.Peek(); que.Dequeue(); // Update prefix_OR prefix_OR |= temp.val; // Update prefix_XOR prefix_XOR ^= temp.val; if (prefix_XOR != prefix_OR) flag = false ; // If left subtree not null if (temp.left != null ) que.Enqueue(temp.left); // If right subtree not null if (temp.right != null ) que.Enqueue(temp.right); // Update length } //If bitwise AND is zero if (flag) uniqueLevels += 1; } Console.Write(uniqueLevels); } // Driver Code public static void Main(String[] args) { TreeNode root = new TreeNode(5); root.left = new TreeNode(6); root.right = new TreeNode(9); root.left.left = new TreeNode(1); root.left.right = new TreeNode(4); root.right.right = new TreeNode(7); // Function Call uniqueLevels(root); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program for the above approach // Structure of a node in // the binary tree class TreeNode { constructor(x) { this .val = x; this .left = null ; this .right = null ; } } // Function to find total unique levels function uniqueLevels(root) { // Stores count of levels, where the set // bits of all the nodes are at // different positions let uniqueLevels = 0; // Store nodes at each level of // the tree using BFS let que = []; que.push(root); // Performing level order traversal while (que.length > 0) { // Stores count of nodes at // current level let length = que.length; // Stores prefix XOR of all // the nodes at current level let prefix_XOR = 0; // Stores prefix OR of all // the nodes at current level let prefix_OR = 0; // Check if set bit of all the nodes // at current level is at different // positions or not let flag = true ; // Traverse nodes at current level for (let i = 0; i < length; i++) { // Stores front element // of the que let temp = que[0]; que.shift(); // Update prefix_OR prefix_OR |= temp.val; // Update prefix_XOR prefix_XOR ^= temp.val; if (prefix_XOR != prefix_OR) flag = false ; // If left subtree not null if (temp.left != null ) que.push(temp.left); // If right subtree not null if (temp.right != null ) que.push(temp.right); } // If bitwise AND is zero if (flag) uniqueLevels += 1; } document.write(uniqueLevels); } // Driver Code let root = new TreeNode(5); root.left = new TreeNode(6); root.right = new TreeNode(9); root.left.left = new TreeNode(1); root.left.right = new TreeNode(4); root.right.right = new TreeNode(7); // Function Call uniqueLevels(root); // This code is contributed by unknown2108 </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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