Question

我制作了一个四叉树，最初是一个统一的2级四叉树。

struct qnode {
  int level;
  double xy[2];
  struct qnode *child[4];
};
typedef struct qnode Node;

int main( int argc, char **argv ) {

  Node *head;

  // make the head node
  head = makeNode( 0.0,0.0, 0 );

  // make a tree

  //full level 2 tree
  makeChildren( head );
  makeChildren( head->child[0] );
  makeChildren( head->child[1] );
  makeChildren( head->child[2] );
  makeChildren( head->child[3] );
}

//在给定位置（x，y）和级别创建一个节点

Node *makeNode( double x, double y, int level ) {

  int i;

  Node *node = (Node *)malloc(sizeof(Node));

  node->level = level;

  node->xy[0] = x;
  node->xy[1] = y;

  for( i=0;i<4;++i )
    node->child[i] = NULL;

  return node;
}

//将叶子节点拆分为4个子对象

void makeChildren( Node *parent ) {

  double x = parent->xy[0];
  double y = parent->xy[1];

  int level = parent->level;

  double hChild = pow(2.0,-(level+1));

  parent->child[0] = makeNode( x,y, level+1 );
  parent->child[1] = makeNode( x+hChild,y, level+1 );
  parent->child[2] = makeNode( x+hChild,y+hChild, level+1 );
  parent->child[3] = makeNode( x,y+hChild, level+1 );

  return;
}

如何遍历每个节点并在每个节点上均匀地生长树？

Answer 1

递归方法似乎是正确的。像这样：

void growTree(Node * root) {
    for(int i=0; i<4; i++) {
        if(root->child[i] == NULL) root->child[i] = makeNode(0,0,root->level+1);
        else growTree(root->child[i]);
    }
}

我还建议重新设计makeNode。像这样：

Node *makeNode(double x, double y, Node * node) {
    int level = node->level;
    // copy old body of makeNode here
}

此外，请更改为此：

typedef struct Node {
  int level;
  double xy[2];
  struct Node *child[4];
} Node;

没有理由为结构节点和节点使用不同的名称。如果要隐藏结构节点的定义，请使用以下方法：

struct Node {
  int level;
  double xy[2];
  struct Node *child[4];
};

typedef struct Node Node;

如何在每个节点上增加四叉树

1 个答案: