I`m slowly implementing some Bitcoin protocols, and I'm not sure how to properly understand the Merkle Tree from the Bitcoin wiki: https://en.bitcoin.it/wiki/Protocol_specification#Merkle_Trees I have strings a, b, c, I double-hash them and I get the three leaf nodes. But I'm not sure what to do with the higher nodes. Do I double hash the sum of the resulting bytes (treating them like big numbers), or do I concatenate (treating them like strings), or perhaps do I should something else?

3 Answers 3


In each iteration, you concatenate two subsequent hashes of the previous level, and double-sha256 them. If there is an odd number of hashes, concatenate the last element with itself. This gives you a new list of hashes. Repeat, and stop when one hash remains. This is the merkle root.

Assume you have tx hashes Ha1,Ha2,Ha3,Ha4,Ha5,Ha6,Ha7

  • First iteration: Hb1=Hash(Ha1|Ha2), Hb2=Hash(Ha3|Ha4), Hb3=Hash(Ha5|Ha6), Hb4=Hash(Ha7|Ha7)
  • Second iteration: Hc1=Hash(Hb1|Hb2), Hc2=Hash(Hb3|Hb4)
  • Third iteration: Hd1=Hash(Hc1|Hc2) => Merkle root

You hash the hashes. See the bottom of https://en.bitcoin.it/wiki/Dump_format#CBlock for an image.


Bitcoin Core's merkle root calculation function can be found here. Here's the simplified version:

// Input: individual transaction hashes (a.k.a. leafs)
uint256 ComputeMerkleRoot(std::vector<uint256> hashes) {
    while (hashes.size() > 1) {
        // If there's an odd number of hashes
        if (hashes.size() & 1) {
            // Copy it to the end of the hashes array
        // Compute double SHA256s of (hashes.size() / 2) blocks
        // Save the output to hashes
        SHA256D64(hashes[0].begin(), hashes[0].begin(), hashes.size() / 2);
        // Save the output to hashes
        hashes.resize(hashes.size() / 2);
    // If there were no original hashes, return 0
    if (hashes.size() == 0) return uint256();
    return hashes[0];

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