PAT程序设计考题——甲级1004 遍历树节点（计算树每层的叶子节点数目）

试题链接如下：点击打开链接

试题要求翻译如下：

程序输入：第一行输入为树总节点个数N（<100）、非叶子节点个数M，用空格分隔；随后输入M行，每行的格式如下：
ID K ID[1] ID[2]...ID[K]
第一个ID代表一个非叶子节点的两位数序号，不满两位用0填充，K代表该节点的子节点数目，随后的K个ID表示子节点的两位数序号。现定root根节点ID是01。
程序输出：算出树每层中叶子节点个数，从根节点所在层开始，逐渐向下层，用空格分隔。
例子：
input：
2 1
01 1 02
output：

0 1

代码设计如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct TreeNode {charvalue;TreeNode**childList;intchildCapacity;intchildNumber;TreeNode( char value ){childNumber = 0;childCapacity = 5;this->value = value;childList = (TreeNode**)malloc(childCapacity*sizeof(TreeNode*));}~TreeNode(){free(childList);childList = NULL;}voidSetNodeValue( char value ){this->value = value;}voidSetNodeList( TreeNode** list ){childList = list;}charGetNodeValue(){return this->value;}TreeNode**GetNodeList(){return childList;}TreeNode*GetNodeListItem(int index){return childList[index];}intAddChild( TreeNode* node ){if ( childNumber == childCapacity ){TreeNode**old = childList;childList = (TreeNode**)realloc( childList, 2*sizeof(TreeNode*)*childCapacity );if ( childList == NULL ){childList = old;return false;}childCapacity = 2*childCapacity;}childList[childNumber++] = node;return true;}}TREE_NODE,*PTREE_NODE;typedef struct LevelInfo {int* number;int  size;int  capacity;LevelInfo(){size = 0;capacity = 5;number = (int*)malloc(capacity*sizeof(int));for ( int i = 0; i < capacity; i++ ){number[i] = 0;}}~LevelInfo(){free(number);}void SetLevelCapacity( int count ){if ( capacity < count ){int* old = number;number = (int*)realloc( number, capacity*2 );if ( number == NULL ){number = old;return;}for ( int i = capacity; i < 2*capacity; i++ ){number[i] = 0;}capacity *= 2;}}void SetLevelSize( int level, int number ){this->number[level] = number;if ( level > size )size = level;}int GetLevelSize( int level ){return number[level];}void ResetAnaly(){for ( int i = 0; i < capacity; i++ ){number[i] = 0;}size = 0;}}LEVEL_INFO,*PLEVELINFO;class CTreeLevelAnaly{public:CTreeLevelAnaly(){InitTree();}~CTreeLevelAnaly(){UninitTree(m_pTreeRoot);delete m_pTreeRoot;}int InitTree(){printf( "Please input the number of node and non-leaf node:\n");scanf( "%d %d", &m_iAllNodeCount, &m_iNonLeafNodeCount );TreeNode** tempList = (TreeNode**)malloc( m_iAllNodeCount*sizeof(TreeNode*) );for ( int i = 0; i < m_iAllNodeCount; i++ ){TreeNode* pNode = new TreeNode( (i+1) );tempList[i] = pNode;}printf( "Please input %d lines describe the parent and chidren\n", m_iNonLeafNodeCount );for ( int k = 0; k < m_iNonLeafNodeCount; k++ ){char* dot = NULL;char buffer[100] = {0};int* list = (int*)malloc( 100*sizeof(int) );int used = 0;printf( "the %d line:\n", k+1 );getchar();scanf( "%[^\n]", buffer );dot = strtok( buffer, " " );while ( dot != NULL ){list[used++] = atoi(dot);dot = strtok( NULL, " ");}list[used] = -1;for ( int j = 2; j < used; j++ ){int index = list[j];int pos = list[0];tempList[ pos - 1 ]->AddChild( tempList[ index - 1 ] );}}m_pTreeRoot = tempList[0];free(tempList);return 0;}voidUninitTree( TreeNode* node ){for (int i = 0; i < node->childNumber; i++ ){UninitTree( node->childList[i] );delete node->childList[i];}}TreeNode*GetRootNode(){return m_pTreeRoot;}int PrintLevelAnaly(){int size = m_kLevelCount.size;for ( int i = 0; i <= size; i++ ){printf("Level %d has %d Leaf-Node\n", i, m_kLevelCount.number[i] );}return false;}// 深度优先，递归 intdfsRecursiveLevelCount( TreeNode* node, int level ){printf("visiting: value:%c;level:%d\n", node->value, level );m_kLevelCount.SetLevelCapacity(level);// 筛选非叶子节点if ( node->childNumber == 0 ){int temp = m_kLevelCount.GetLevelSize( level );if ( temp == 0 ){temp = 1;}else{temp++;}m_kLevelCount.SetLevelSize( level, temp );}for (int i = 0; i < node->childNumber; i++ ){dfsRecursiveLevelCount( node->childList[i], level + 1 );}return 0;}protected:private:TreeNode*m_pTreeRoot;LevelInfom_kLevelCount;intm_iAllNodeCount;intm_iNonLeafNodeCount;};int main(){CTreeLevelAnaly* tree = new CTreeLevelAnaly;printf("-----------深度优先，递归------------\n");tree->dfsRecursiveLevelCount( tree->GetRootNode(), 0 );tree->PrintLevelAnaly();delete tree;printf("\n");return 0;}