Question

这是一个找到二叉搜索树的最低共同祖先的函数。它工作正常，因为它在函数中正确打印LCA但在主函数中它只有在LCA等于root否则返回值时才有效是NULL。可以解释一下吗？

node* FindLCA(node* root,int val1,int val2)
  {
 if(root->val<val1&&root->val>val2||root->val>val1&&root->val<val2||root->val==val1||root->v    al==val2)
    { cout<<"\n\n"<<"LCA of "<<val1<<"and "<<val2<<"is "<<root->val<<"\n";  //working correctly here
    return root;
    }

 if(root->val>val1&&root->val>val2)
    FindLCA(root->left,val1,val2);
 else
    FindLCA(root->right,val1,val2);
 }

 int main()
 {
  int arr[]={5,2,6,1,7,4,8,9};
  node* tree=buildtree(arr,8);
  printPretty((BinaryTree*)tree, 1, 0, cout);
  int a=1,b=2;
  node* lca=FindLCA(tree,a,b);
if(lca!=NULL) cout<<"\n\n"<<"LCA of "<<a<<"and "<<b<<"is "<<lca->val<<"\n";            //working only if LCA equals root's value
 system("pause");
  }

Answer 1

您应该return FindLCA(root->right,val1,val2);而不是仅仅呼叫它。您只在root时返回一个值，这就是为什么在主要输出正确的情况下它是唯一的情况。

node* FindLCA(node* root,int val1,int val2)
{
  if((root->val<val1&&root->val>val2) || (root->val>val1&&root->val<val2) ||
      root->val==val1 || root->val==val2)
  { 
    cout<<"\n\n"<<"LCA of "<<val1<<"and "<<val2<<"is "<<root->val<<"\n";  //working correctly here
    return root;
  }

  if(root->val>val1&&root->val>val2)
    return FindLCA(root->left,val1,val2);   // RETURN VALUE HERE
  else
    return FindLCA(root->right,val1,val2);  // RETURN VALUE HERE
}

int main()
{
  int arr[]={5,2,6,1,7,4,8,9};
  node* tree=buildtree(arr,8);
  printPretty((BinaryTree*)tree, 1, 0, cout);
  int a=1,b=2;
  node* lca=FindLCA(tree,a,b);
  if(lca!=NULL) cout<<"\n\n"<<"LCA of "<<a<<"and "<<b<<"is "<<lca->val<<"\n";  //working only if LCA equals root's value
  system("pause");
}

BST的最低共同祖先（返回值中的陌生行为）

1 个答案: