【编程之美】3.10分层遍历二叉树

主要难度在于何时插入换行

学习到的：①vector 可以像数组一样用 不一定要用迭代器

 

代码及注释如下：

#include<iostream>

#include<queue>

using namespace std;



typedef struct BiTree

{

    BiTree * pLeft, * pRight;

    int data;

}BiTree;



void createBiTree(BiTree * &T)

{

    int d;

    cout << "please input the data of tree:";

    cin >> d;

    if(d != 0)

    {

        T = new BiTree;

        T->data = d;

        T->pLeft = NULL;

        T->pRight = NULL;

        createBiTree(T->pLeft);

        createBiTree(T->pRight);

    }

}

//分层打印整个二叉树 答案的思路 用游标cur last标示回车位置 不用辅助结点

void PrintNodeByLevel(BiTree* root)

{

    if(root == NULL)

    {

        return;

    }

    vector<BiTree *> vec;

    vec.push_back(root);

    int cur = 0;

    int last = 1;

    while(cur < vec.size())

    {

        last = vec.size(); //新一行访问开始 重定位last于当前行最后一个结点的下一个结点

        while(cur < last)

        {

            printf("%d ", vec[cur]->data); //注意 vector也可以像数组一样用

            if(vec[cur]->pLeft != NULL)

            {

                vec.push_back(vec[cur]->pLeft);

            }

            if(vec[cur]->pRight != NULL)

            {

                vec.push_back(vec[cur]->pRight);

            }

            cur++;

        }

        printf("\n");

    }

}



//分层打印整个二叉树 我自己的思路 利用数据为‘\n’的辅助的树节点标示回车位置

void Travese(BiTree * T)

{

    queue<BiTree *> Q;



    BiTree Assist;

    Assist.data = '\n';  //辅助表示回车

    Assist.pLeft = NULL;

    Assist.pRight = NULL;



    Q.push(T);

    Q.push(&Assist);

    while(!Q.empty())

    {

        BiTree * TmpTree = Q.front();

        Q.pop();

        if(TmpTree->data != '\n')

        {

            if(TmpTree->pLeft != NULL) //注意  空指针不压入 因为下面还要判断其下一个有效树结点是不是回车

            {

                Q.push(TmpTree->pLeft);

            }

            if(TmpTree->pRight != NULL)

            {

                Q.push(TmpTree->pRight);

            }

            if(Q.front()->data == '\n') //当一个\n要被弹出时 压入下一个\n

            {

                Q.push(&Assist);

            }

            printf("%d ", TmpTree->data);

            

        }

        else if(TmpTree->data == '\n')

        {

            printf("\n");

        }

    }

}



//打印二叉树中指定层的结点 根节点为第0层

int PrintNodeAtLevel(BiTree* root, int level)

{

    if(root == NULL || level < 0)

    {

        return 0;

    }

    if(level == 0)

    {

        printf("%d ", root->data);

        return 1;

    }

    else

    {

        int returnl = 1, returnr = 1;

        if(root->pLeft != NULL)

        {

            returnl = PrintNodeAtLevel(root->pLeft, level - 1);

        }

        if(root->pRight != NULL)

        {

            returnr = PrintNodeAtLevel(root->pRight, level - 1);

        }

        return (returnl && returnr) ? 1 : 0;

    }

}

int main()

{

    BiTree * T = NULL;

    createBiTree(T);

    Travese(T);

    PrintNodeAtLevel(T, 2);

    printf("\n");

    PrintNodeByLevel(T);

    return 0;

}