1
问题二叉树节点:我需要能够递归通过索引检索未知高度的完美二叉树一个节点。递归指数通过其广度优先指数
由于未知height属性的似乎是有道理的索引的唯一形式是广度优先索引(按照标题):
0
1 2
3 4 5 6
的问题是,在每一个节点似乎难以知道要采取哪个方向,以及如何将递归请求中的索引转换为该子节点......或者我只是没有清楚地思考。
Node Navigate(Index):
Index 0: return this;
Index 1: return Left.Navigate(0);
Index 2: return Right.Navigate(0);
Index 3: return Left.Navigate(1);
Index 4: return Left.Navigate(2);
Index 5: return Right.Navigate(1);
Index 6: return Right.Navigate(2);
...
Index 7: return Left.Navigate(3);
Index 8: return Left.Navigate(4);
Index 9: return Left.Navigate(5);
Index 10: return Left.Navigate(6);
Index 11: return Right.Navigate(3);
Index 12: return Right.Navigate(4);
Index 13: return Right.Navigate(5);
Index 14: return Right.Navigate(6);
模式是明确的 - 但如何能一个编程 - 无需消耗太多的时钟周期(这是一种嵌入装置) - 从Index
选择一个节点和变换Index
到参数用于导航该节点?我错过了一个简单的转换?
这是我结束了使用,建立在yurib的答案实现:
public class Node
{
public Node Left, Right;
public Node(int levels)
{
if (levels == 0) return;
Left = new Node(levels - 1);
Right = new Node(levels - 1);
}
public Node Navigate(int index)
{
if (index == 0) return this;
// we want 1 based indexing.
int oneBased = index + 1;
// level is how many levels deep we are looking, stored as 1 << depth.
int level = 1;
// find level - it's equal to the highest bit in "oneBased". Find it.
for (int i = oneBased; (i >>= 1) != 0;)
{
level *= 2;
}
// level adjusted for our children.
int subLevel = level >> 1;
// clear our level bit, set our children's level bit.
int childIndex = ((oneBased & ~level) | subLevel) - 1;
// is the node we're looking for over half way through level? go right.
if ((oneBased & subLevel) != 0)
return Right.Navigate(childIndex);
else
return Left.Navigate(childIndex); // otherwise it's in our left tree.
}
}
这是C#进行测试,但在现实中每个呼叫进行浏览不同的嵌入式设备上进行处理,因此需要为递归而不是简单地跟随伪代码,建立一个List
等感谢yurib :)。
精彩的回答! – noMAD 2012-02-28 06:34:26