二叉树 - 计算级别上的节点

Question

我正在写一个二叉树类,而且我停留在一个levelCount方法,我需要计算树级别上的节点数.类和方法看起来像这样:

public class ConsTree<T> extends BinaryTree<T>
{
   BinaryTree<T> left;
   BinaryTree<T> right;
   T data;

   public int levelCount(int level) 
   {
   }
}  

所以我们的想法是每棵树左边都有一棵树,右边有一棵树,还有数据.有一个抽象类binarytree和子类ConsTree和EmptyTree.

我想我需要使用广度优先搜索并在达到该级别后计算节点数量,但我仍然坚持如何开始.这里的任何指导都会有所帮助.我可以提供任何其他必要的信息.

Answer 1

这是一般方法.

您可以像平常一样遍历树(深度优先,按顺序),但您只需传递所需的和实际的级别,使用伪代码,例如:

def getCountAtLevel (node, curr, desired):
    # If this node doesn't exist, must be zero.
    if node == NULL: return 0

    # If this node is at desired level, must be one.
    if curr == desired: return 1

    # Otherwise sum of nodes at that level in left and right sub-trees.
    return getCountAtLevel (node.left,  curr+1, desired) +
           getCountAtLevel (node.right, curr+1, desired)

#######################################################################
# Get number of nodes at level 7 (root is level 0).
nodesAtLevel7 = getCountAtLevel (rootNode, 0, 7)

它实际上并不遍历整个树,因为一旦达到所需的级别,它就可以忽略它下面的所有内容.这是一个完整的C程序,它显示了这一点:

#include <stdio.h>

typedef struct _sNode { struct _sNode *left, *right; } tNode;

// Node naming uses (t)op, (l)eft, and (r)ight.
tNode TLLL = {NULL,  NULL    }; // level 3
tNode TLLR = {NULL,  NULL    };
tNode TRLL = {NULL,  NULL    };
tNode TRLR = {NULL,  NULL    };
tNode TRRR = {NULL,  NULL    };
tNode TLL  = {&TLLL, &TLLR   }; // level 2
tNode TRL  = {&TRLL, &TRLR   };
tNode TRR  = {NULL,  &TRRR   };
tNode TL   = {&TLL,  NULL    }; // level 1
tNode TR   = {&TRL,  &TRR    };
tNode T    = {&TL,   &TR     }; // level 0 (root)

static int getCAL (tNode *node, int curr, int desired) {
    if (node == NULL) return 0;
    if (curr == desired) return 1;
    return getCAL (node->left,  curr+1, desired) +
           getCAL (node->right, curr+1, desired);
}

int main (void) {
    for (int i = 0; i < 5; i++) {
        int count = getCAL(&T, 0, i);
        printf ("Level %d has %d node%s\n", i, count, (count == 1) ? "" : "s");
    }
    return 0;
}

它构建了以下形式的树:

            ______T______               (1 node)
           /             \
         TL               TR            (2 nodes)
        /                /  \
     TLL              TRL    TRR        (3 nodes)
    /   \            /   \      \
TLLL     TLLR    TRLL     TRLR   TRRR   (5 nodes)

                                        (0 nodes)

然后为您提供每个级别的节点数:

Level 0 has 1 node
Level 1 has 2 nodes
Level 2 has 3 nodes
Level 3 has 5 nodes
Level 4 has 0 nodes