如何在右子树上找到最小的数字

I want to find the smallest number in a rignt subtree of a node, and this code below is what I thought that was the solution, but its not working properly. What is wrong with this code?

我想在一个节点的严格子树中找到最小的数字，下面这段代码就是我认为的解决方案，但它无法正常工作。这段代码有什么问题？

int small; // Where the smallest value is stored

int smallest(Node n)
{
    if(n.info < small && aux != 0) small = n.info;
    if(aux == 0)
    {
        aux = 1;
        small = n.dir.info;
        if(n!=NULL && n.dir!=NULL) return smallest(n.dir);
    }
    else{
        if(n.dir != NULL) return smallest(n.dir);
        if(n.esq != NULL) return smallest(n.esq); 
    }
    return small;
}

1 个解决方案

#1

I am using n.right for right subtree pointer and n.left for left subtree pointer

我使用n.right作为右子树指针，n.left用于左子树指针

Just call the function smallest(n.right); smallest is a function that will find the smallest value in a binary tree

只需调用函数最小（n.right）; smallest是一个可以在二叉树中找到最小值的函数

int smallest(Node n){
  if( n==NULL ) return INF;  // replace INF with the maximum value that int can hold in your system like 2147483647
  int left_small = smallest(n.left); // smallest value in left subtree
  int right_small = smallest(n.right); // smallest value in right subtree
  int ans = n.info;
  if( left_small < ans ) ans = left_small;
  if( right_small < ans ) ans = right_small;
  return ans;
}

#1