/*
This implementation demonstrates how
to efficiently find the diameter of a
binary tree. Note that the diameter
of a binary tree is defined as the
length of its longest path.
Let n be the number of nodes and h be
the height of the binary tree.
Time complexity: O(n)
Space complexity: O(h)
*/
public class BinaryTreeDiameter {
BTNode BTRoot;
public BinaryTreeDiameter() {
/*
* Create tree below:
* * 1
* /
* 2 3
* * /
* * 4 5
*/
BTRoot = new BTNode(1, null, null);
BTNode rootLeft = new BTNode(2, null, null);
BTRoot.left = rootLeft;
BTNode rootRight = new BTNode(3, null, null);
BTRoot.right = rootRight;
BTNode rootRightLeft = new BTNode(4, null, null);
BTNode rootRightRight = new BTNode(5, null, null);
rootRight.left = rootRightLeft;
rootRight.right = rootRightRight;
}
public static void main(String[] args) {
BinaryTreeDiameter application = new BinaryTreeDiameter();
System.out.println(application.binaryTreeDiameter(application.BTRoot)); // 3
}
public int binaryTreeDiameter(BTNode root) {
int[] diameter = new int[1];
findDiameter(root, diameter);
return diameter[0];
}
private int findDiameter(BTNode node, int[] diameter) {
if (node == null) {
return 0;
}
int leftHeight = findDiameter(node.left, diameter);
int rightHeight = findDiameter(node.right, diameter);
diameter[0] = Math.max(diameter[0], leftHeight + rightHeight);
return 1 + Math.max(leftHeight, rightHeight);
}
// Class representing a binary tree node
// with pointers to value, left, right, and parent nodes
private class BTNode {
int val;
BTNode left;
BTNode right;
public BTNode(int val, BTNode left, BTNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}