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）及运行实例结果

#include<iostream>
#define N 7
using 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
={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);
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}
}
}

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;
}