Question

让结构给出：

// Struct to nAry tree
struct nNode {
  int val;                  // Some value (in future use a pointer to some data)
  struct nNode *next;       // Siblings of the same parent if next == NULL is the last
  struct nNode *prev;       // Siblings of same parent if prev == NULL is the first
  struct nNode *parent;     // Points to parent, if NULL is the root 
  struct nNode *children;   // Child node, other childs are found by moving next/prev
                            // if NULL is a leaf node
};

下面的代码应该给出Traverse in Level

void nNode_traverse_levelOrder(struct nNode *node)
{
  struct nNode *child;
  struct nNode *sibling;
  struct nNode *head;

  struct QueueLL *queue;
  queue = newQueueLL();

  // Queue the root node
  enqueueLL(queue, node);

  while (! isQueueEmptyLL(queue)) {
    head = dequeueLL(queue);

    if(head) {
      visit(head);

      sibling = head->next;
      // Queue all brothers
      while(sibling) {
        enqueueLL(queue, sibling);
        sibling = sibling->next;
      }

      // Queue the children (there is only one)
      child = head->children;
      if (child) 
        enqueueLL(queue, child);
    }
  }
  destroyQueueLL(queue);
  queue = NULL;
}

鉴于树：

  /*                      1
   *            /---------|--------\
   *          2           3         4
   *        /   \                 /
   *      5       6              7
   */

返回

Node val: 1
Node val: 2
Node val: 3
Node val: 4
Node val: 5
Node val: 4
Node val: 7
Node val: 6
Node val: 7

但预期是1 2 3 4 5 6 7.我已经使用Pre，Pos和In顺序遍历函数进行了双重检查，所有这些都正确返回，如下所示：

PRE 1 2 _5 _6 _3 4 _7

POS _5 _6 2 _3 _7 4 1

IN _5 2 _6 1 _3 _7 4

想弄清楚在我的功能中可能误导我的是什么

Answer 1

当您访问节点时，您将跟随所有兄弟姐妹。下一个兄弟姐妹会再次将其余的兄弟姐妹排队。如果您有一个可以在队列开头添加元素的操作，则可以在对子进行排队后使用：

if (sibling) { pushLL(queue, sibling); }

或者只是将孩子排队，而不是兄弟姐妹，这是通常的方式，并使功能更短：

void nNode_traverse_levelOrder(struct nNode* node) { struct QueueLL* queue = newQueueLL(); enqueueLL(queue, node); while (!isQueueEmptyLL(queue)) { struct nNode* head = dequeueLL(queue); visit(head); for (struct nNode* child = head->children; child != NULL; child = child->next) { enqueueLL(queue, child); } } destroyQueueLL(queue); }

遍历n-Ary树级别顺序

1 个答案: