Check whether a given binary tree is perfect or not
Given a Binary Tree, write a function to check whether the given Binary Tree is a prefect Binary Tree or not.
A Binary tree is Perfect Binary Tree in which all internal nodes have two children and all leaves are at same level.
Examples:
The following tree is a perfect binary tree
10
/ \
20 30
/ \ / \
40 50 60 70
18
/ \
15 30
The following tree is not a perfect binary tree
1
/ \
2 3
\ / \
4 5 6
A Perfect Binary Tree of height h (where height is number of nodes on path from root to leaf) has 2h – 1 nodes.
Below is an idea to check whether a given Binary Tree is perfect or not.
- Find depth of any node (in below tree we find depth of leftmost node). Let this depth be d.
- Now recursively traverse the tree and check for following two conditions.
- Every internal node should have both children non-empty
- All leaves are at depth ‘d’
C++
// C++ program to check whether a given// Binary Tree is Perfect or not#include<bits/stdc++.h> /* Tree node structure */struct Node{ int key; struct Node *left, *right;}; // Returns depth of leftmost leaf.int findADepth(Node *node){ int d = 0; while (node != NULL) { d++; node = node->left; } return d;} /* This function tests if a binary tree is perfect or not. It basically checks for two things : 1) All leaves are at same level 2) All internal nodes have two children */bool isPerfectRec(struct Node* root, int d, int level = 0){ // An empty tree is perfect if (root == NULL) return true; // If leaf node, then its depth must be same as // depth of all other leaves. if (root->left == NULL && root->right == NULL) return (d == level+1); // If internal node and one child is empty if (root->left == NULL || root->right == NULL) return false; // Left and right subtrees must be perfect. return isPerfectRec(root->left, d, level+1) && isPerfectRec(root->right, d, level+1);} // Wrapper over isPerfectRec()bool isPerfect(Node *root){ int d = findADepth(root); return isPerfectRec(root, d);} /* Helper function that allocates a new node with the given key and NULL left and right pointer. */struct Node *newNode(int k){ struct Node *node = new Node; node->key = k; node->right = node->left = NULL; return node;} // Driver Programint main(){ struct Node* root = NULL; root = newNode(10); root->left = newNode(20); root->right = newNode(30); root->left->left = newNode(40); root->left->right = newNode(50); root->right->left = newNode(60); root->right->right = newNode(70); if (isPerfect(root)) printf("Yes\n"); else printf("No\n"); return(0);} |
Java
// Java program to check whether a given // Binary Tree is Perfect or not class GfG { /* Tree node structure */static class Node { int key; Node left, right; } // Returns depth of leftmost leaf. static int findADepth(Node node) { int d = 0; while (node != null) { d++; node = node.left; } return d; } /* This function tests if a binary tree is perfect or not. It basically checks for two things : 1) All leaves are at same level 2) All internal nodes have two children */static boolean isPerfectRec(Node root, int d, int level) { // An empty tree is perfect if (root == null) return true; // If leaf node, then its depth must be same as // depth of all other leaves. if (root.left == null && root.right == null) return (d == level+1); // If internal node and one child is empty if (root.left == null || root.right == null) return false; // Left and right subtrees must be perfect. return isPerfectRec(root.left, d, level+1) && isPerfectRec(root.right, d, level+1); } // Wrapper over isPerfectRec() static boolean isPerfect(Node root) { int d = findADepth(root); return isPerfectRec(root, d, 0); } /* Helper function that allocates a new node with the given key and NULL left and right pointer. */static Node newNode(int k) { Node node = new Node(); node.key = k; node.right = null; node.left = null; return node; } // Driver Program public static void main(String args[]) { Node root = null; root = newNode(10); root.left = newNode(20); root.right = newNode(30); root.left.left = newNode(40); root.left.right = newNode(50); root.right.left = newNode(60); root.right.right = newNode(70); if (isPerfect(root) == true) System.out.println("Yes"); else System.out.println("No"); }} |
Python3
# Python3 program to check whether a # given Binary Tree is Perfect or not # Helper class that allocates a new # node with the given key and None # left and right pointer. class newNode: def __init__(self, k): self.key = k self.right = self.left = None # Returns depth of leftmost leaf. def findADepth(node): d = 0 while (node != None): d += 1 node = node.left return d # This function tests if a binary tree# is perfect or not. It basically checks # for two things : # 1) All leaves are at same level # 2) All internal nodes have two children def isPerfectRec(root, d, level = 0): # An empty tree is perfect if (root == None): return True # If leaf node, then its depth must # be same as depth of all other leaves. if (root.left == None and root.right == None): return (d == level + 1) # If internal node and one child is empty if (root.left == None or root.right == None): return False # Left and right subtrees must be perfect. return (isPerfectRec(root.left, d, level + 1) and isPerfectRec(root.right, d, level + 1)) # Wrapper over isPerfectRec() def isPerfect(root): d = findADepth(root) return isPerfectRec(root, d) # Driver Code if __name__ == '__main__': root = None root = newNode(10) root.left = newNode(20) root.right = newNode(30) root.left.left = newNode(40) root.left.right = newNode(50) root.right.left = newNode(60) root.right.right = newNode(70) if (isPerfect(root)): print("Yes") else: print("No") # This code is contributed by pranchalK |
C#
// C# program to check whether a given // Binary Tree is Perfect or not using System; class GfG { /* Tree node structure */class Node { public int key; public Node left, right; } // Returns depth of leftmost leaf. static int findADepth(Node node) { int d = 0; while (node != null) { d++; node = node.left; } return d; } /* This function tests if a binary tree is perfect or not. It basically checks for two things : 1) All leaves are at same level 2) All internal nodes have two children */static bool isPerfectRec(Node root, int d, int level) { // An empty tree is perfect if (root == null) return true; // If leaf node, then its depth must be same as // depth of all other leaves. if (root.left == null && root.right == null) return (d == level+1); // If internal node and one child is empty if (root.left == null || root.right == null) return false; // Left and right subtrees must be perfect. return isPerfectRec(root.left, d, level+1) && isPerfectRec(root.right, d, level+1); } // Wrapper over isPerfectRec() static bool isPerfect(Node root) { int d = findADepth(root); return isPerfectRec(root, d, 0); } /* Helper function that allocates a new node with the given key and NULL left and right pointer. */static Node newNode(int k) { Node node = new Node(); node.key = k; node.right = null; node.left = null; return node; } // Driver code public static void Main() { Node root = null; root = newNode(10); root.left = newNode(20); root.right = newNode(30); root.left.left = newNode(40); root.left.right = newNode(50); root.right.left = newNode(60); root.right.right = newNode(70); if (isPerfect(root) == true) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } /* This code is contributed by Rajput-Ji*/ |
Output:
Yes
Time complexity : O(n)
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