algorithm - 只有左 child 的二叉树中的节点数

Question

如何找到只有左 child 的二叉树中的节点数？

LeftNode(root)
{
    if(root==NULL) 
        return 0;

    if(root->left!=null && root-right==null)
        return (1+LeftNode(root->left));

    return (LeftNode (root->left) + LeftNode (root->right))
}

Answer 1

我会这样做(C++):

int leftNode(Node * root)
{
  if (root == nullptr)
    return 0;

  // c is 1 if root has a left child and has not a right one
  int c = root->left != nullptr and root->right == nullptr ? 1 : 0;

  return c + leftNode(root->left) + leftNode(root->right);
}