Question

详细的问题是从给定节点找到距离为x（即边数= x）的所有节点。

我今天在亚马逊采访中被问到，

void findNodeWithDistanceX(struct node* root,struct node * qnode, int value)
{
    //root is root Node, qnode is questionnode from which distance to be calculated, value is the            
    //distance to be calculated


    //finding distance between root and qnode
    int distance = findDistancefromRoot(root ,qnode);

    if(distance> value)
    {
        traverseDistancedown(root ,distance-value);
    }

    if(distance ==value){
        printf("%d",root->value);
    }

    // Traverse and find all nodes with distance value from 'qnode' down the tree
    traverseDistancedown(qnode,value);

现在如果找到距离＆＃34;值＆＃34;来自qnode 我没有得到如何遍历树并满足距离值的条件的答案。

虽然这里出现了很多案例。

我尝试从qnode回溯，但无济于事我无法给他留下深刻印象，无法编写代码。

关于在树上实现遍历的任何讨论对我都有帮助。

Answer 1

最简单的方法是使用深度优先搜索，随时跟踪距离。它是一个简单的预订遍历，从＆＃34;问题节点＆＃34;开始。

void nodesAtDistance(struct node* qnode, int requestedDistance, int currentDistance)
{
    if (qnode == null) return;

    if (currentDistance == requestedDistance)
    {
        // output qnode
        // and then return because children are further away
        return;
    }

    // visit left and then right
    nodesAtDistance(qnode->left, requestedDistance, currentDistance+1);
    nodesAtDistance(qnode->right, requestedDistance, currentDistance+1);
}

所以如果你用：

来称呼它

nodesAtDistance(myNode, 5, 0);

它将输出低于myNode 5级的所有节点。

现在，如果我将此作为API函数编写，我将提供一个不需要客户端传递currentDistance参数的重载。因为如果有人写nodesAtDistance(myNode, 5, 1)，你就会得到距离4的东西。所以，创造这个过载：

void nodesAtDistance(node * qnode, int requestedDistance)
{
    nodesAtDistance(qnode, requestedDistance, 0);
}

将该功能公开显示，并将另一个功能设为私有。

Answer 2

尝试此代码，虽然此代码仅检查子节点，但不会返回检查给定距离的祖先节点

    #include<stdio.h>
    #include<malloc.h>

    struct treenode
    {
     unsigned int data;
     struct treenode *left;
     struct treenode *right;
    };

    struct treenode  *treeptr, *sourcenode, *quesnode, *quesleftnode, *quesrightnode, *root =          NULL;

    struct node *treeptr insert_node(unsigned int treedata)
    {
      treeptr= (struct node*)(malloc(sizeof(treenode)));
      treeptr->data = nodevalue;
      treeptr->left   = NULL;
      treeptr->right = NULL;    
      return treeptr;
   }

   printnodesdistancek(struct treenode* quesnode, unsigned char reqdist)
   {
     unsigned char curdist=0;
     do
     {
      if(quesnode == null)
      return;

      quesleftnode = quesnode->left;

       if(curdist == reqdist)
       {
         printf("%d",quesleftnode->data);
       } 
       else
       {
        quesnode= quesnode->left;
       }
       quesrightnode = quesnode->right;

      if(curdist == reqdist)
      {
        printf("%d",quesrightnode->data);
      }
      else
      {
       quesnode = quesnode->right;
      }
      }while(quesnode!=NULL);
      //main function
      void main(void)
      {
       //create tree

       *root = insert_node(1);
        root->left=insert_node(-1);
        root->right=insert_node(2);
        root->left->left=insert_node(-2);
        root->left->right=insert_node(3);
        root->right->left=insert_node(-3);
        root->right->right=insert_node(4);
        sourcenode = root->left;
        printnodesdistancek(sourcenode,1);

      }

打印距离为＆＃39; x＆＃39;的所有节点来自BST中的给定节点

2 个答案: