m

```
package com.thealgorithms.datastructures.trees;
import java.util.HashMap;
import java.util.Stack;
/**
* This class will check if a BinaryTree is balanced. A balanced binary tree is
* defined as a binary tree where the difference in height between the left and
* right subtree of each node differs by at most one.
* <p>
* This can be done in both an iterative and recursive fashion. Below,
* `isBalancedRecursive()` is implemented in a recursive fashion, and
* `isBalancedIterative()` is implemented in an iterative fashion.
*
* @author [Ian Cowan](<a href="https://github.com/iccowan">Git-Ian Cowan</a>)
*/
public class CheckIfBinaryTreeBalanced {
/**
* Recursive is BT balanced implementation
*
* @param root The binary tree to check if balanced
*/
public static boolean isBalancedRecursive(BinaryTree.Node root) {
if (root == null) {
return true;
}
// Create an array of length 1 to keep track of our balance
// Default to true. We use an array, so we have an efficient mutable object
boolean[] isBalanced = new boolean[1];
isBalanced[0] = true;
// Check for balance and return whether we are balanced
isBalancedRecursive(root, 0, isBalanced);
return isBalanced[0];
}
/**
* Private helper method to keep track of the depth and balance during
* recursion. We effectively perform a modified post-order traversal where
* we are looking at the heights of both children of each node in the tree
*
* @param node The current node to explore
* @param depth The current depth of the node
* @param isBalanced The array of length 1 keeping track of our balance
*/
private static int isBalancedRecursive(BinaryTree.Node node, int depth, boolean[] isBalanced) {
// If the node is null, we should not explore it and the height is 0
// If the tree is already not balanced, might as well stop because we
// can't make it balanced now!
if (node == null || !isBalanced[0]) {
return 0;
}
// Visit the left and right children, incrementing their depths by 1
int leftHeight = isBalancedRecursive(node.left, depth + 1, isBalanced);
int rightHeight = isBalancedRecursive(node.right, depth + 1, isBalanced);
// If the height of either of the left or right subtrees differ by more
// than 1, we cannot be balanced
if (Math.abs(leftHeight - rightHeight) > 1) {
isBalanced[0] = false;
}
// The height of our tree is the maximum of the heights of the left
// and right subtrees plus one
return Math.max(leftHeight, rightHeight) + 1;
}
/**
* Iterative is BT balanced implementation
*/
public static boolean isBalancedIterative(BinaryTree.Node root) {
if (root == null) {
return true;
}
// Default that we are balanced and our algo will prove it wrong
boolean isBalanced = true;
// Create a stack for our post order traversal
Stack<BinaryTree.Node> nodeStack = new Stack<>();
// For post order traversal, we'll have to keep track of where we
// visited last
BinaryTree.Node lastVisited = null;
// Create a HashMap to keep track of the subtree heights for each node
HashMap<BinaryTree.Node, Integer> subtreeHeights = new HashMap<>();
// We begin at the root of the tree
BinaryTree.Node node = root;
// We loop while:
// - the node stack is empty and the node we explore is null
// AND
// - the tree is still balanced
while (!(nodeStack.isEmpty() && node == null) && isBalanced) {
// If the node is not null, we push it to the stack and continue
// to the left
if (node != null) {
nodeStack.push(node);
node = node.left;
// Once we hit a node that is null, we are as deep as we can go
// to the left
} else {
// Find the last node we put on the stack
node = nodeStack.peek();
// If the right child of the node has either been visited or
// is null, we visit this node
if (node.right == null || node.right == lastVisited) {
// We assume the left and right heights are 0
int leftHeight = 0;
int rightHeight = 0;
// If the right and left children are not null, we must
// have already explored them and have a height
// for them so let's get that
if (node.left != null) {
leftHeight = subtreeHeights.get(node.left);
}
if (node.right != null) {
rightHeight = subtreeHeights.get(node.right);
}
// If the difference in the height of the right subtree
// and left subtree differs by more than 1, we cannot be
// balanced
if (Math.abs(rightHeight - leftHeight) > 1) {
isBalanced = false;
}
// The height of the subtree containing this node is the
// max of the left and right subtree heights plus 1
subtreeHeights.put(node, Math.max(rightHeight, leftHeight) + 1);
// We've now visited this node, so we pop it from the stack
nodeStack.pop();
lastVisited = node;
// Current visiting node is now null
node = null;
// If the right child node of this node has not been visited
// and is not null, we need to get that child node on the stack
} else {
node = node.right;
}
}
}
// Return whether the tree is balanced
return isBalanced;
}
}
```