Find the Distance between Two Nodes of a Binary Tree.

Objective: – Given nodes in a binary tree, find the distance between them.

Example :

Approach:

• Distance(X, Y) = Distance(root, X) +Distance(root, Y) – 2*(Distance(root to LCA(X,Y)
• where LCA(X,Y) = Lowest Common Ancestor of X,Y
• In the above example if Distance(20,45) = 3
• Distance(root, 20) = 2
• Distance(root, 45) = 3
• LCA(20, 45) = 10
• Distance(root, 10) = 1
• Distance(20,45) = 2+3 – 2*1 = 3

Complete Code:

	public class PrintDistance {
	public int findDistance(Node root, int n1, int n2) {
	int x = Pathlength(root, n1) - 1;
	int y = Pathlength(root, n2) - 1;
	int lcaData = findLCA(root, n1, n2).data;
	int lcaDistance = Pathlength(root, lcaData) - 1;
	return (x + y) - 2 * lcaDistance;
	}
	public int Pathlength(Node root, int n1) {
	if (root != null) {
	int x = 0;
	if ((root.data == n1) || (x = Pathlength(root.left, n1)) > 0
	|| (x = Pathlength(root.right, n1)) > 0) {
	// System.out.println(root.data);
	return x + 1;
	}
	return 0;
	}
	return 0;
	}
	public Node findLCA(Node root, int n1, int n2) {
	if (root != null) {
	if (root.data == n1 || root.data == n2) {
	return root;
	}
	Node left = findLCA(root.left, n1, n2);
	Node right = findLCA(root.right, n1, n2);
	if (left != null && right != null) {
	return root;
	}
	if (left != null) {
	return left;
	}
	if (right != null) {
	return right;
	}
	}
	return null;
	}
	public static void main(String[] args) throws java.lang.Exception {
	Node root = new Node(5);
	root.left = new Node(10);
	root.right = new Node(15);
	root.left.left = new Node(20);
	root.left.right = new Node(25);
	root.right.left = new Node(30);
	root.right.right = new Node(35);
	root.left.right.right = new Node(45);
	PrintDistance i = new PrintDistance();
	System.out.println("Distance between 45 and 20 is : "
	+ i.findDistance(root, 45, 20));
	}
	}
	class Node {
	int data;
	Node left;
	Node right;
	public Node(int data) {
	this.data = data;
	this.left = null;
	this.right = null;
	}
	}