二叉树-在一个级别上计数节点 [英] Binary Tree - Counting nodes on a level

问题描述

我正在编写一个二叉树类，并且被困在levelCount方法上，该方法需要计算树级别上的节点数.该类和方法如下所示:

I'm writing a binary tree class, and I'm stuck on a levelCount method, where I need to count the number of nodes on a level of the tree. The class and method look something like this:

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

   public int levelCount(int level) 
   {
   }
}  

因此，想法是每棵树在其左侧都有一棵树，在其右侧有一棵树以及数据.有一个抽象类binarytree和子类ConsTree和EmptyTree.

So the idea is that each tree has a tree to its left, a tree to its right, and data. There is an abstract class binarytree and subclasses ConsTree and EmptyTree.

我认为我需要使用广度优先搜索并在达到该级别后计算节点数，但是我对如何开始一无所知.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.

I think I need to use a breadth first search and count the number of nodes once I get to that level, but I'm stuck on how to start. Any guidance here would be helpful. I can provide any other info necessary.

推荐答案

以下是一般方法.

您可以像平常一样遍历树(深度优先，顺序排列)，但是您也可以通过伪代码简单地传递所需的和实际的级别，例如:

You traverse the tree exactly as you would normally (depth-first, in-order) but you simply pass down the desired and actual level as well, with pseudo-code such as:

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)

它实际上并没有遍历 entire 树，因为一旦到达所需的级别，它就可以忽略其中的所有内容.这是一个完整的C程序，展示了这一点:

It doesn't actually traverse the entire tree since, once it gets to the desired level, it can just ignore everything underneath that. Here's a complete C program that shows this in action:

#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;
}

它将构建以下形式的树(其中T表示顶部，L是左分支，而R是右分支):

It builds a tree of the following form (where T means top, L is a left branch and R is a right branch):

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

                                        (0 nodes)

如果编译并运行该代码，您会看到它在每个级别提供了正确的节点数:

If you compile and run that code, you'll see it gives the correct node count at each level:

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

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