Before discussing the current problem let me give brief introduction of binary trees and the basic interface and classes Iam using for coding Diameter of a Binary tree.

**Binary Tree**

A binary tree is made of nodes, where each node contains a “left” pointer, a “right” pointer, in additional a “parent” pointer and a data element. The “root” pointer points to the topmost node in the tree. The left and right pointers recursively point to smaller “subtrees” on either side. A null pointer represents a binary tree with no elements — the empty tree.

The formal recursive definition is: a **binary tree **is either empty (represented by a null pointer), or is made of a single node, where the left and right pointers (recursive definition ahead) each point to a **binary tree**.

Class for Node is as follows:

package ds; public class Node { Node left; Node right; Node parent; int data; }

Interface for a Tree is as follows:

package ds; public interface Tree { public Node search(int data); }

Class for Binary tree which has create method, add method, search method and print method is as follows:

package ds; import java.util.LinkedList; public class BinaryTree implements Tree { public static final String LEFT = "LEFT"; public static final String RIGHT = "RIGHT"; private Node root; public BinaryTree() { } public void create(int currentData) { if (root == null) { root = new Node(); } root.data = currentData; } public void add(Node node, int currentData, String currentPosition) { if (node == null) { return; } Node currentNode = new Node(); currentNode.data = currentData; if (LEFT.equals(currentPosition)) { node.left = currentNode; } else if (RIGHT.equals(currentPosition)) { node.right = currentNode; } } public Node search(int searchData) { if (root == null) { return null; } return search(searchData, root); } private Node search(int searchData, Node node) { if (node == null) { return null; } // Level order traversal to return the node LinkedList queueList = new LinkedList(); queueList.add(node); Node currNode = null; while (!queueList.isEmpty()) { currNode = (Node) queueList.removeFirst(); if (currNode.data == searchData) { break; } else { if (currNode.left != null) { queueList.add(currNode.left); } if (currNode.right != null) { queueList.add(currNode.right); } } } return currNode; } public void printLevelOrder() { printLevelOrder(root); System.out.println(); } public void printLevelOrder(Node node) { if (node == null) { return; } LinkedList queueList = new LinkedList(); queueList.add(node); Node currNode = null; while (!queueList.isEmpty()) { currNode = (Node) queueList.removeFirst(); System.out.print(currNode.data + " "); if (currNode.left != null) { queueList.add(currNode.left); } if (currNode.right != null) { queueList.add(currNode.right); } } System.out.println(); } public void printInorder() { printInorder(root); System.out.println(); } public void printInorder(Node node) { if(node == null) { return; } printInorder(node.left); System.out.print(node.data+" "); printInorder(node.right); } public void printPreorder() { printPreorder(root); System.out.println(); } public void printPreorder(Node node) { if(node == null) { return; } System.out.print(node.data+" "); printPreorder(node.left); printPreorder(node.right); } public void printPostorder() { printPostorder(root); System.out.println(); } public void printPostorder(Node node) { if(node == null) { return; } printPostorder(node.left); printPostorder(node.right); System.out.print(node.data+" "); } }

Diameter of a binary tree

Diameter of a Tree is defined as the number of nodes on the longest path between two leaves in the tree.

Have a look at the following figures:

In the above first figure the longest path between the leaves is through the root. And in the second figure the longest path between the leaves is in the left subtree of the root.

Hence the diameter of a binary tree T will be the largest of the following quantities:

- the diameter of
*T*’s left subtree - the diameter of
*T*’s right subtree - the longest path between leaves that goes through the root of
*T*(this can be computed from the heights of the subtrees of*T*)

Dry run for the following example:

Consider following binary tree:

At every node its left height, right height, left diameter and right diameter are presented:

Therefore for the root node, diameter is maximum of left diameter: 5, right diameter: 3, left height+ right height+1: 4+2+1. Here its 7.

The code for the above problem is as follows:

package ds; public class DiameterOfTree { public DiameterOfTree() { } public int diameterOfBinaryTree(Node node) { if (node == null) { return 0; } int leftHeight = heightOfBinaryTree(node.left); int rightHeight = heightOfBinaryTree(node.right); int leftDiameter = diameterOfBinaryTree(node.left); int rightDiameter = diameterOfBinaryTree(node.right); return Math.max(leftHeight + rightHeight + 1, Math.max(leftDiameter, rightDiameter)); } public int heightOfBinaryTree(Node node) { if (node == null) { return 0; } else { return 1 + Math.max(heightOfBinaryTree(node.left), heightOfBinaryTree(node.right)); } } public static void main(String[] args) { /* Constructing following tree 1 2 3 4 5 6 7 8 9 10 11 */ BinaryTree binaryTree = new BinaryTree(); binaryTree.create(1); Node root = binaryTree.search(1); binaryTree.add(root, 2, BinaryTree.LEFT); binaryTree.add(root, 3, BinaryTree.RIGHT); binaryTree.add(binaryTree.search(2), 4, BinaryTree.LEFT); binaryTree.add(binaryTree.search(2), 5, BinaryTree.RIGHT); binaryTree.add(binaryTree.search(3), 6, BinaryTree.LEFT); binaryTree.add(binaryTree.search(3), 7, BinaryTree.RIGHT); binaryTree.add(binaryTree.search(4), 8, BinaryTree.LEFT); binaryTree.add(binaryTree.search(4), 9, BinaryTree.RIGHT); binaryTree.add(binaryTree.search(9), 10, BinaryTree.LEFT); binaryTree.add(binaryTree.search(9), 11, BinaryTree.RIGHT); System.out.println("Binary Tree in level order is .... "); binaryTree.printLevelOrder(); DiameterOfTree diaTree = new DiameterOfTree(); System.out.println("Diameter is .... " + diaTree.diameterOfBinaryTree(root)); } }

References

http://www.cs.duke.edu/courses/spring00/cps100/assign/trees/diameter.html

Please write comments if you find any of the above algorithms or codes incorrect, or find better ways to solve the same problem.

[…] Implementation for Node, Tree, BinaryTree can be obtained from my previous post. […]

[…] Implementation for Node, Tree, BinaryTree can be obtained from my previous post. […]

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Shouldn’t line 22:

… leftHeight + rightHeight + 1,

be:

… lh + rh + 2,

? After all, there’s an edge, counting for 1, from us to both the left and right child nodes, for a total of 2. And we’re counting the dist from the tip of one to the tip of the other. It’s not a mere +1 for our own, single-level of additional height.

Also, take eg leftHeight. You’re not distinguishing between the left child having no children (returning height 0), and /there being no left child/. You need to do the null test before calling height, and not add 1 if there is no child. The same of course for the right child.

Hi, Yes, I thought so too. But look at the definition of diameter. It is the number of nodes, not number of edges.

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I think that instead of leftHeight + rightHeight + 1 there should be leftHeight + rightHeight. Correct me if I’m wrong

If we assume the height at node level is 1, then it should be (leftHeight + rightHeight – 1). Please let me know if I’m wrong.

height at root level is 1, not node.

While I agree with your solution, I am having hard find figuring out why my solution might be incorrect. Appreciate your comments.

As I see, we can maintain a max_sum variable and count the height of the left subtree (lh), height of right subtree (rh) and update the max_sum if lh + rh + 1 is its greater. We calculate this sum at each of the nodes in the tree.

From your solution, ultimately the diameters will be calculated based on heights of the left and right subtrees at some point of time in the code.

As I see the mentioned algo is correct but takes O(n^2). I suppose O(n) algorithm is possible here. Just modify the main function to calculate both diameter and height.

There are three cases to consider when trying to find the longest path between two nodes in a binary tree (diameter):

The longest path passes through the root,

The longest path is entirely contained in the left sub-tree,

The longest path is entirely contained in the right sub-tree.

The longest path through the root is simply the sum of the heights of the left and right sub-trees + 1 (for the root node), and the other two can be found recursively:

public static int getDiameter(BinaryTreeNode root) {

if (root == null)

return 0;

int rootDiameter = getHeight(root.getLeft()) + getHeight(root.getRight()) + 1;

int leftDiameter = getDiameter(root.getLeft());

int rightDiameter = getDiameter(root.getRight());

return Math.max(rootDiameter, Math.max(leftDiameter, rightDiameter));

}

public static int getHeight(BinaryTreeNode root) {

if (root == null)

return 0;

return Math.max(getHeight(root.getLeft()), getHeight(root.getRight())) + 1;

}

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The article is good read, I slightly doubt its efficiency. However, with a choice of slightly intelligent data structure we can achieve linear time complexity in an iterative approach. Check that out. I would have loved if you could explain the calculation of time complexity.

http://techieme.in/tree-diameter/

Here is the simple algorithm to find width of a binary tree:

* Find height of the binary tree, let it be H.

* Traverse the tree using pre order traversal and find number of nodes in one level at a time.

* Check if the number of nodes in current level is more than the maximum width we found till now. If yes then update maximum width of binary tree.

* In this approach, we will traverse given tree H-1 times because in every traversal we are finding node count of one leve

Here is the C program link http://www.techcrashcourse.com/2016/06/program-to-find-maximum-width-of-binary-tree.html

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