Question

我将下面的代码写到BST中 count 个大于给定KEY的节点数：

int Tree::findNumbersThatBiggerThanKey(int key){
    return PrivateFindNumbersThatBiggerThanKey(key, this->rootNode);
}

int Tree::PrivateFindNumbersThatBiggerThanKey(int key, Node* start) {
    if (!start)
        return 0;

    if (start->key > key)
        return 1 + this->PrivateFindNumbersThatBiggerThanKey(key, start->leftNode) +
        this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

    else if (start->key < key)
        return this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

    else /*start->key > key*/
        return this->PrivateFindNumbersThatBiggerThanKey(key, start->leftNode);
}

这是我的主要爱好

int main() {

    int TreeKeys[16] = { 50, 76, 21, 4, 32, 64, 15, 52, 14, 100, 83, 2, 3, 70, 87, 80 };
    Tree myBst;

    cout << "Printing the three Inorder before adding numbers: \n";
    myBst.printInorder();


    for (int i = 0; i < 16; i++) {
        myBst.AddLeaf(TreeKeys[i]);
    }

    cout << "Printing the three Inorder after adding numbers: \n";
    myBst.printInorder();

    int key = 2;
    cout << "\n\nNumber of nodes that are bigger than "<<key<<" : " <<
         myBst.findNumbersThatBiggerThanKey(key);
    return 0;
}

Inroder打印效果很好：

Printing the three Inorder after adding numbers:
2 3 4 14 15 21 32 50 52 64 70 76 80 83 87 100

但是当我搜索大于“ 2”的节点数时，我得到14个错误的值（而不是15个）：

Number of nodes that are bigger than 2 : 14

当我搜索大于“ 1”的节点数时，我得到16个正确的结果（对于其他任何数字我也得到正确的结果）：

Number of nodes that are bigger than 1 : 16

我是递归的学生和新手，我很高兴为您解释为什么发生这种情况以及如何解决它。

Answer 1

您的else不正确，它的条件是start->key == key，应与start->key < key一样对待

因此将其重写为

else 
   return this->PrivateFindNumbersThatBiggerThanKey(key, start->rightNode);

或仅将其与start->key < key

组合

Answer 2

由于您未提供MCVE，因此我无法处理您的代码。这样做的一般方法是这样的：

int Tree::PrivateFindNumbersThatBiggerThanKey(int key, Node* node)
{
  if (node == null)
  {
    return 0;
  }
  int countLeft = PrivateFindNumbersThatBiggerThanKey(node->leftNode, key);
  int countRight = PrivateFindNumbersThatBiggerThanKey(node->rightNode, key);

  return (node->key > k ? 1 : 0) + countLeft + countRight;
}

BST中大于给定KEY的节点数

2 个答案: