## Balanced Binary Search Trees

### March 3, 2017

The story exploded across the intertubes a few days ago: A software engineer trying to enter the United States on a work visa was asked to prove his occupation by writing a program for the Customs and Border Protection agent:

Write a function to check if a Binary Search Tree is balanced.

Let’s help him.

Your task is to write a function to check if a binary search tree is balanced. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Untested, of course. I’m trying to simulate my being in the situation of that engineer :(

[sourcelang=”css”]

def makeNode(left=None, right=None):

(left, right)

def left(node):

node[0]

def right(node):

node[1]

left = makeNode(left=makeNode(left=makeNode(left=makeNode(left=makeNode()))))

print(“{} is {}balanced.”.format(‘left’,”” if isbalanced(left) else “not “))

left is balanced.

# good job Python!

[/sourcelang=”css”]

In Common Lisp:

Python functions need to have return statements. They return None by default.

[…] a recent exercise we wrote a program to determine if a binary tree was balanced. Today’s exercise is […]

A C++ solution that checks for near perfect balance.

[sourcelang=”c++”]

struct Node

{

int key;

Node * left, * right;

Node(int k)

{

key = k;

left = right = NULL;

}

};

int height(Node * node)

{

if(node == NULL)

return 0;

return max( height(node->left), height(node->right));

}

bool IsBalanced(Node * node)

{

int diff = height(node->left) – height(node->right);

return ( -1 <= diff && diff left) && IsBalanced(node->right));

}

[/sourcelang=”c++”]

Here’s a solution in Java. It uses pre-order traversal, and it each node checks that the heights of the left sub-tree and right sub-tree are within 1. The height calculation uses memoization to avoid redundant calculations.

Output: