Stern-Brocot Tree

Stern-Brocot tree is a tree data structure whose vertices correspond to the set of non-negative rational numbers. Thus, this tree provides a very elegant way for constructing the set of fractions m/n, where m and n are relatively prime. To construct the tree, the basic idea is to start with two fractions (0/1, 1/0) and then repeat the following operation:

Insert (m+m’)/(n+n’) between two adjacent fractions m/n and m’/n’

The first step gives us the entry 1/1 between 0/1 and 1/0. Similarly, the 2nd step gives us two more: 0/1,1/2,1/1,2/1,1/0

Continuing on like this results in an infinite binary search tree which preserves the usual ordering of rational numbers.

The figure below shows the 1st 4 levels of the Stern-Brocot tree.


Finding the Path to k in Stern-Brocot Tree

The path from the root of the tree to a number k in the Stern-Brocot tree can be found using binary search. At each node, k will either be in the left half of the tree, or the right half. We continue down the left or right subtree until we finally find k.

  • Initialize the left fraction L to 0/1 and right fraction R to 1/0
  • Repeat the following until k is found:
    • Compute the mediant M (which is (m+m’)/(n+n’) )
    • If (k>M), then k is in the right half of the tree. L:=M and continue.
    • Else If (M>k), then k is in the left half of the tree. R:=M and continue.
    • Else k=M, terminate search.


There’s a couple of things to tackle in our implementation. First, I need an easy way to represent fractions, so I create my own SternBrocotFraction class. I deliberately chose to make it very specific to this algorithm because I needed a special way to handle the fraction 1/0 (which by definition is greater than all other rationals).

Secondly, I needed a good way to represent the path from the root of the tree to k. I do this by using a StringBuilder, and at each step I append either the letter ‘L’ or ‘R’ depending on which sub-tree we take. When the search is finished, this gives us a String representation of the path from the root of the tree to the number k. This approach is similar to the approach advocated by ACM Programming Competitions for the “Stern-Brocot Number System” problem.

Here’s the code to find path to a number k:

package com.umairsaeed.algorithm;

public class SternBrocotPath {
	private static final char LEFT_SUB = 'L';
	private static final char RIGHT_SUB = 'R';

	public String findPathTo(SternBrocotFraction f) {
		SternBrocotFraction L = new SternBrocotFraction(0, 1);
		SternBrocotFraction R = new SternBrocotFraction(1, 0);

		StringBuilder results = new StringBuilder();
		SternBrocotPath.find(f, L, R, results);
		return results.toString();

	public static void find(SternBrocotFraction f,
			SternBrocotFraction L,
			SternBrocotFraction R,
			StringBuilder results)
		SternBrocotFraction M = L.add(R);

		if (M.compareTo(f) < 0) {
			L = M;
			SternBrocotPath.find(f, L, R, results);
		} else if (M.compareTo(f) > 0) {
			R = M;
			SternBrocotPath.find(f, L, R, results);

The special SternBrocotFraction class is:

package com.umairsaeed.algorithm;

public class SternBrocotFraction implements
				Comparable<SternBrocotFraction> {
	private int numerator;
	private int denominator;

	public SternBrocotFraction(int numerator, int denominator) {
		if (denominator < 0) {
			numerator *= -1;
			denominator *= -1;

		this.numerator = numerator;
		this.denominator = denominator;

	public double doubleValue() {
		if (this.denominator == 0) {
			return Double.MAX_VALUE;
		} else {
			return (double) this.numerator /
						(double) this.denominator;

	public SternBrocotFraction add(SternBrocotFraction other) {
		return new SternBrocotFraction(
				this.numerator + other.numerator,
				this.denominator + other.denominator);

	public int compareTo(SternBrocotFraction other) {
		if (this.doubleValue() < other.doubleValue()) {
			return -1;
		} else if (this.doubleValue() > other.doubleValue()) {
			return 1;
		return 0;

Finally, some test code to exercise my class:

package com.umairsaeed.algorithm;

public class SternBrocotTester {
	public static void main(String[] args) {

	public static void testSternBrocotPath() {
		SternBrocotPath t = new SternBrocotPath();

		SternBrocotFraction f = new SternBrocotFraction(5, 7);

		f = new SternBrocotFraction(19, 101);

		f = new SternBrocotFraction(977, 331);

		f = new SternBrocotFraction(1049, 7901);
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Filed under Algorithms

2 responses to “Stern-Brocot Tree

  1. pedant

    Else q=M, terminate search.


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