c++实现二叉树的查找，插入，删除，深度，叶子节点数，度为1的节点数（递归方法）

这是算法导论中二叉树搜索的一个例题

二叉树为

c++代码为

#include<iostream>#define N 7using namespace std;//二叉树节点类class node{public:    int data;    node *leftChild;    node *rightChild;};typedef node *BiTree;//等价//创建节点node *createNode(int value){    node *q=new node;    q->leftChild=NULL;    q->rightChild=NULL;    q->data=value;    return q;}//创建二叉树BiTree createBiTree(){    node *p[N]={NULL};    int array[6]={6,5,7,2,5,8};    for(int i=0;i<6;++i)        p[i]=createNode(array[i]);    for(int i=0;i<N/2;++i)    {        p[i]->leftChild=p[i*2+1];        p[i]->rightChild=p[i*2+2];    }    return p[0];//返回根节点}//求二叉树的深度int depth(BiTree tree){    int dep=0;    int leftDep,rightDep;    if(!tree)        dep=0;    else    {        leftDep=depth(tree->leftChild);        rightDep=depth(tree->rightChild);        dep=1+(leftDep>rightDep?leftDep:rightDep);    }    return dep;}//度为1的节点数,即只有一个分支int numOfOneDgree(BiTree tree){    int sum=0;    int leftNum,rightNum;    if(tree)    {        if(((tree->leftChild!=NULL)&&(tree->rightChild==NULL))||((tree->leftChild==NULL)&&(tree->rightChild!=NULL)))            sum++;        leftNum=numOfOneDgree(tree->leftChild);//递归        sum+=leftNum;        rightNum=numOfOneDgree(tree->rightChild);        sum+=rightNum;    }    return sum;}//叶子节点的个数int sumOfSubLeft(BiTree tree){    int sum=0;    int leftNum,rightNum;    if(tree)    {        if((!tree->leftChild)&&(!tree->rightChild))//左右节点均为NULL            sum++;        leftNum=sumOfSubLeft(tree->leftChild);        sum+=leftNum;        rightNum=sumOfSubLeft(tree->rightChild);        sum+=rightNum;    }    return sum;}//查找bool search(BiTree tree,int searchNum){    if(tree==NULL)    {        return 0;    }    else    {        if(tree->data==searchNum)        {            return 1;        }        else        {            if(searchNum<tree->data)                return search(tree->leftChild,searchNum);            else                return search(tree->rightChild,searchNum);        }    }}//访问节点中的数据int visit(BiTree tree){    return tree->data;}// 中序遍历void inorderTreeWalk(BiTree tree){    if(tree)    {        inorderTreeWalk(tree->leftChild);        cout << visit(tree) << " ";        inorderTreeWalk(tree->rightChild);    }}//插入void insertNode(node* *tree ,int x){    if(*tree==NULL)//新建一个根节点    {        node *p=new node;        p->data=x;        p->leftChild=NULL;        p->rightChild=NULL;        *tree=p;        return;    }    else if(x<(*tree)->data)        insertNode(&((*tree)->leftChild),x);    else        insertNode(&((*tree)->rightChild),x);}//删除,比较麻烦,分几种情况int deleteNode(node* *tree,int x){    node *temp=*tree;    if(*tree==NULL)        return 0;    if(x<(*tree)->data)        return deleteNode(&((*tree)->leftChild),x);    if(x>(*tree)->data)        return deleteNode(&((*tree)->rightChild),x);    if((*tree)->leftChild==NULL)//左子树为空，且删除元素等于根节点，把右子树作为整个树    {        *tree=(*tree)->rightChild;        free(temp);        return 1;    }    else    {        if((*tree)->leftChild->rightChild==NULL)//前驱节点为空时，把左孩子节点赋给根节点，再从左子树中删除根节点            {                (*tree)->data=(*tree)->leftChild->data;                return deleteNode(&((*tree)->leftChild),(*tree)->data);            }        else        {//找到前驱节点，再把该节点赋给根节点，最后删除该根节点            node *p1=*tree;            node *p2=p1->leftChild;            while(p2->rightChild!=NULL)            {                p1=p2;                p2=p2->rightChild;            }            (*tree)->data=p2->data;            return deleteNode(&(p1->rightChild),p2->data);        }    }}int main(){    BiTree tree;    tree=createBiTree();    cout<<"二叉树的深度为："<<depth(tree)<<endl;    cout<<"二叉树的叶子节点个数为："<<sumOfSubLeft(tree)<<endl;    cout<<"二叉树中度为1的个数为："<<numOfOneDgree(tree)<<endl;    cout<<"请输入要查找的数,输入1000结束查找"<<endl;    int n,m,w;    while(1)    {        cin>>n;        if(n==1000)            break;        else        {            if(search(tree,n))                cout<<"查找到值为"<<n<<"的数"<<endl;            else                cout<<"未查到到值为"<<n<<"的数"<<endl;        }    }    cout << endl<< "中序遍历结果为:";    inorderTreeWalk(tree);    cout<<endl<<"请输入要插入的值:";    cin>>m;    insertNode(&tree,m);    cout << "插入值后中序遍历结果为:";    inorderTreeWalk(tree);    cout<<endl<<"请输入要删除的值:";    cin>>w;    deleteNode(&tree,w);    cout << "删除值后中序遍历结果为:";    inorderTreeWalk(tree);    return 0;}

运行结果为