# Check whether a given binary tree is perfect or not

• Difficulty Level : Easy
• Last Updated : 27 Jun, 2022

Given a Binary Tree, write a function to check whether the given Binary Tree is a perfect 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.

1. Find depth of any node (in below tree we find depth of leftmost node). Let this depth be d.
2. 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`   `/*  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 Program` `int` `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*/`

## Javascript

 ``

Output:

`Yes`

Complexity Analysis:

Time complexity: O(n)

Space Complexity: O(n)

Method 2: Using the length of the binary tree

Since a full binary tree has 2^h – 1 nodes, we can count the number of nodes in the binary tree and determine whether it is a power of 2 or not.

To efficiently determine whether it is a power of 2 or not, we can use bitwise operation x & (x+1) == 0

## C++

 `// C++ program to check whether a` `// given Binary Tree is Perfect or not` `#include ` `using` `namespace` `std;`   `// Helper class that allocates a new` `// node with the given key and None` `// left and right pointer.` `class` `newNode{` `    ``public``:` `    ``int` `key;` `    ``newNode*right,*left;` `    ``newNode(``int` `k){` `        ``this``->key = k;` `        ``this``->right = ``this``->left = NULL;` `    ``}` `};`   `// This functions gets the size of binary tree` `// Basically, the number of nodes this binary tree has` `int` `getLength(newNode* root){` `    ``if``(root == NULL)` `        ``return` `0;` `    ``return` `1 + getLength(root->left) + getLength(root->right);` `}`   `// Returns True if length of binary tree is a power of 2 else False` `bool` `isPerfect(newNode* root){` `    ``int` `length = getLength(root);` `    ``return` `length & (length+1) == 0;` `}`   `int` `main()` `{` `    ``newNode* root = ``new` `newNode(10);` `    ``root->left = ``new` `newNode(20);` `    ``root->right = ``new` `newNode(30);`   `    ``root->left->left = ``new` `newNode(40);` `    ``root->left->right = ``new` `newNode(50);` `    ``root->right->left = ``new` `newNode(60);` `    ``root->right->right = ``new` `newNode(70);`   `    ``if` `(isPerfect(root))` `        ``cout<<``"Yes"``<

## Java

 `/*package whatever //do not write package name here */`   `import` `java.io.*;`   `class` `GFG ` `{` `  `  `// Java 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.` `static` `class` `newNode{` `    ``public` `int` `key;` `    ``public` `newNode right,left;` `    ``public` `newNode(``int` `k){` `        ``this``.key = k;` `        ``this``.right = ``this``.left = ``null``;` `    ``}` `};`   `// This functions gets the size of binary tree` `// Basically, the number of nodes this binary tree has` `static` `int` `getLength(newNode root){` `    ``if``(root == ``null``)` `        ``return` `0``;` `    ``return` `1` `+ getLength(root.left) + getLength(root.right);` `}`   `// Returns True if length of binary tree is a power of 2 else False` `static` `boolean` `isPerfect(newNode root){` `    ``int` `length = getLength(root);` `    ``return` `(length & (length+``1``))== ``0``;` `}`   `/* Driver program to test above function*/` `public` `static` `void` `main(String args[])` `{` `    ``newNode root = ``new` `newNode(``10``);` `    ``root.left = ``new` `newNode(``20``);` `    ``root.right = ``new` `newNode(``30``);`   `    ``root.left.left = ``new` `newNode(``40``);` `    ``root.left.right = ``new` `newNode(``50``);` `    ``root.right.left = ``new` `newNode(``60``);` `    ``root.right.right = ``new` `newNode(``70``);`   `    ``if` `(isPerfect(root))` `        ``System.out.println(``"Yes"``);` `    ``else` `        ``System.out.println(``"No"``);` `}` `}`   `// This code is contributed by shinjanpatra`

## 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`   `#This functions gets the size of binary tree` `#Basically, the number of nodes this binary tree has` `def` `getLength(root):` `  ``if` `root ``=``=` `None``:` `    ``return` `0` `  ``return` `1` `+` `getLength(root.left) ``+` `getLength(root.right)`   `#Returns True if length of binary tree is a power of 2 else False` `def` `isPerfect(root):` `  ``length ``=` `getLength(root)` `  ``return` `length & (length``+``1``) ``=``=` `0`   `# 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 beardedowl`

## Javascript

 ``

Complexity Analysis:

Time Complexity: O(n)

Space Complexity:O(n)
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