C++实现二叉树的相应操作

2021-07-12 04:06

阅读:801

标签:二叉树的遍历   turn   tree node   amp   scom   main   二叉树   com   技术   

1. 二叉树的遍历:先序(递归、非递归),中序(递归、非递归),后序(递归、非递归)。

#include 
#include string>
#include using namespace std;

struct BiTree
{
    int NodeData = 0;
    struct BiTree *pLeft = nullptr;
    struct BiTree *pRight = nullptr;
};

//遍历二叉树:
void show(struct BiTree *pRoot, int n)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {
        show(pRoot->pLeft, n + 1);

        for (int i = 0; i )
            cout "   ";
        cout NodeData  endl;

        show(pRoot->pRight, n + 1);
    }

}
//--------------------------------------------------------------
//递归中序遍历:
void RecMidTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else  
    {
        if (pRoot->pLeft != nullptr)
        {
            RecMidTravel(pRoot->pLeft);
        }

        cout NodeData  endl;

        if (pRoot->pRight != nullptr)
        {
            RecMidTravel(pRoot->pRight);
        }
    }
} 

//中序非递归
void MidTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {

        struct BiTree *pcur = pRoot;
        stack mystack;

        while (!mystack.empty() || pcur != nullptr)
        {
            while (pcur != nullptr)
            {
                mystack.push(pcur);
                pcur = pcur->pLeft;    //左节点全部进栈
            }

            if (!mystack.empty())
            {
                pcur = mystack.top();
                cout NodeData  endl;
                mystack.pop();                    //出栈
                pcur = pcur->pRight;            //右节点
            }
        }

    }
}
//--------------------------------------------------------------
//递归先序遍历:
void RecPreTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {
        cout NodeData  endl;

        if (pRoot->pLeft != nullptr)
        {
            RecPreTravel(pRoot->pLeft);
        }

        if (pRoot->pRight != nullptr)
        {
            RecPreTravel(pRoot->pRight);
        }
    }
}
//先序非递归
void PreTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {

        struct BiTree *pcur = pRoot;
        stack mystack;

        while (!mystack.empty() || pcur != nullptr)
        {
            while (pcur != nullptr)
            {
                cout NodeData  endl;

                mystack.push(pcur);
                pcur = pcur->pLeft;    //左节点全部进栈
            }

            if (!mystack.empty())
            {
                pcur = mystack.top();
                
                mystack.pop();                    //出栈
                pcur = pcur->pRight;            //右节点
            }
        }

    }
}

//--------------------------------------------------------------
//递归后序遍历:
void RecPostTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {
        if (pRoot->pLeft != nullptr)
        {
            RecPostTravel(pRoot->pLeft);
        }

        if (pRoot->pRight != nullptr)
        {
            RecPostTravel(pRoot->pRight);
        }

        cout NodeData  endl;
    }
}
//后序非递归
struct nosame    //标识节点是否反复出现
{
    struct BiTree *pnode;
    bool issame;
};

void PostTravel(struct BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return;
    }
    else
    {

        struct BiTree *pcur = pRoot;
        stack mystack;    //避免重复出现
        nosame *ptemp;

        while (!mystack.empty() || pcur != nullptr)
        {
            while (pcur != nullptr)
            {
                nosame *ptr = new nosame;
                ptr->issame = true;
                ptr->pnode = pcur;//节点


                //cout NodeData 

                mystack.push(ptr);
                pcur = pcur->pLeft;    //左节点全部进栈
            }

            if (!mystack.empty())
            {
                ptemp = mystack.top();
                mystack.pop();                    //出栈

                if (ptemp->issame == true)        //第一次出现
                {
                    ptemp->issame = false;
                    mystack.push(ptemp);
                    pcur = ptemp->pnode->pRight;//跳到右节点
                }
                else
                {
                    cout pnode->NodeData //打印数据
                    pcur = nullptr;
                }

            }
        }

    }
}


void main()
{
    struct BiTree *pRoot;

    struct BiTree node1;
    struct BiTree node2;
    struct BiTree node3;
    struct BiTree node4;
    struct BiTree node5;
    struct BiTree node6;
    struct BiTree node7;
    struct BiTree node8;

    node1.NodeData = 1;
    node2.NodeData = 2;
    node3.NodeData = 3;
    node4.NodeData = 4;
    node5.NodeData = 5;
    node6.NodeData = 6;
    node7.NodeData = 7;
    node8.NodeData = 8;

    pRoot = &node1;
    node1.pLeft = &node2;
    node1.pRight = &node3;

    node2.pLeft = &node4;
    node2.pRight = &node5;

    node3.pLeft = &node6;
    node3.pRight = &node7;

    node4.pLeft = &node8;

    show(pRoot, 1);

    cout "中序递归:"  endl;
    RecMidTravel(pRoot);    //中序递归
    cout "中序非递归:"  endl;
    MidTravel(pRoot);        //中序非递归

    cout "先序递归:"  endl;
    RecPreTravel(pRoot);
    cout "先序非递归:"  endl;
    PreTravel(pRoot);        //先序非递归

    cout "后序递归:"  endl;
    RecPostTravel(pRoot);
    cout "后序非递归:"  endl;
    PostTravel(pRoot);        //后序非递归


    cin.get();
}

     技术分享图片

 2. 获取二叉树节点个数:

//递归获取二叉树节点个数
int getNodeCount(BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return 0;
    }
    else
    {
        return getNodeCount(pRoot->pLeft) + getNodeCount(pRoot->pRight) + 1;
    }
}

    技术分享图片

3. 判断二叉树是否为完全二叉树:

//判断二叉树是否为完全二叉树
bool isCompleteBiTree(BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        return false;
    }
    else
    {
        queue myq;
        myq.push(pRoot);
        bool mustHaveChild = false;    //必须有子节点
        bool result = true;            //结果

        while (!myq.empty())
        {
            BiTree *node = myq.front();//头结点
            myq.pop();                    //出队

            if (mustHaveChild)    //必须有孩子
            {
                if (node->pLeft != nullptr || node->pRight != nullptr)
                {
                    result = false;
                    break;

                }

            } 
            else
            {
                if (node->pLeft != nullptr && node->pRight != nullptr)
                {
                    myq.push(node->pLeft);
                    myq.push(node->pRight);
                }
                else if (node->pLeft != nullptr && node->pRight == nullptr)
                {
                    mustHaveChild = true;
                    myq.push(node->pLeft);
                }
                else if (node->pLeft == nullptr && node->pRight != nullptr)
                {
                    result = false;
                    break;
                }
                else
                {
                    mustHaveChild = true;
                }
            }
        }

        return result;
    }
}

    技术分享图片

4. 求二叉树两个节点的最小公共祖先:

//求二叉树两个节点的最小公共祖先
bool findnode(BiTree *pRoot, BiTree *node)    //判断节点是否在某个节点下
{
    if (pRoot == nullptr || node == nullptr)
    {
        return false;
    }
    if (pRoot == node)
    {
        return true;
    }

    bool isfind = findnode(pRoot->pLeft, node);
    if (!isfind)
    {
        isfind = findnode(pRoot->pRight, node);
    }

    return isfind;
}

BiTree *getParent(BiTree *pRoot, BiTree *pChild1, BiTree *pChild2)
{
    if (pRoot == pChild1 || pRoot == pChild2)
    {
        return pRoot;
    }

    if (findnode(pRoot->pLeft, pChild1))
    {
        if (findnode(pRoot->pRight, pChild2))
        {
            return pRoot;
        } 
        else
        {
            return getParent(pRoot->pLeft, pChild1, pChild2);
        }
    } 
    else
    {
        if (findnode(pRoot->pLeft, pChild2))
        {
            return pRoot;
        }
        else
        {
            return getParent(pRoot->pRight, pChild1, pChild2);
        }
    }
}

    技术分享图片

5. 二叉树的翻转:

//二叉树的翻转
BiTree *revBiTree(BiTree *pRoot)
{
    if (pRoot==nullptr)
    {
        return nullptr;
    }

    BiTree *leftp = revBiTree(pRoot->pLeft);
    BiTree *rightp = revBiTree(pRoot->pRight);

    pRoot->pLeft = rightp;
    pRoot->pRight = leftp;    //交换

    return pRoot;
}

    技术分享图片

6. 求二叉树第k层的节点个数:

//求二叉树第K层的节点个数
int getLevelConut(BiTree *pRoot, int k)
{
    if (pRoot == nullptr || k 1)
    {
        return 0;
    }
    if (k == 1)
    {
        return 1;
    }
    else
    {
        int left = getLevelConut(pRoot->pLeft, k - 1);
        int right = getLevelConut(pRoot->pRight, k - 1);

        return (left + right);
    }
}

    技术分享图片

7. 求二叉树中节点的最大距离(相距最远的两个节点之间的距离):

//求二叉树中节点的最大距离
struct res    //用以递归间传递距离
{
    int maxDistance = 0;
    int maxDepth = 0;
};

res getMaxDistance(BiTree *pRoot)
{
    if (pRoot == nullptr)
    {
        res r1;
        return r1;
    }

    res leftr = getMaxDistance(pRoot->pLeft);
    res rightr = getMaxDistance(pRoot->pRight);

    res last;    //最终结果
    last.maxDepth = max(leftr.maxDepth + 1, rightr.maxDepth + 1);//求最大深度
    last.maxDistance = max(max(leftr.maxDistance, rightr.maxDistance), leftr.maxDepth + rightr.maxDepth + 2);//求最大距离

    return last;
}

    技术分享图片

8. 判断二叉树是否为平衡二叉树:

//判断二叉树是否为平衡二叉树:
bool isAVL(BiTree *pRoot, int & depth)    //需要引用来传递数据
{
    if (pRoot == nullptr)
    {
        depth = 0;
        return true;
    }

    int leftdepth = 0;
    int rightdepth = 0;
    bool left = isAVL(pRoot->pLeft, leftdepth);
    bool right = isAVL(pRoot->pRight, rightdepth);

    if (left && right && abs(leftdepth - rightdepth) 1)
    {
        depth = 1 + (leftdepth > rightdepth ? leftdepth : rightdepth);//深度
        return true;
    }
    else
    {
        return false;
    }
}

    技术分享图片    技术分享图片

 

C++实现二叉树的相应操作

标签:二叉树的遍历   turn   tree node   amp   scom   main   二叉树   com   技术   

原文地址:https://www.cnblogs.com/si-lei/p/9547016.html


评论


亲,登录后才可以留言!