【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;
}