今天去面试了，有2道题没做出来，发现自己做题真的不行啊，晚上恶补。

1 二叉树分层遍历

二叉树遍历，要求从上往下遍历每一层，在每一层里从左往右遍历。题目里提示用队列，一直没想出来队列怎么用，后来自己想一下，直接用2个链表就搞定了，一个链表存储上一层结点，另外一个链表存储下一层结点。

#include<stdio.h>
#include<stdlib.h>
#include<assert.h>

typedef struct BinaryTreeNode BTNode;
struct BinaryTreeNode
{
    int value;
    BTNode *pLeft;
    BTNode *pRight;
};

typedef struct List List;
struct List
{
    BTNode *pNode;
    List *pNext;
};

BTNode* newNode(int value)
{
    BTNode *p = (BTNode*)malloc(sizeof(BTNode));
    p->value = value;
    p->pLeft = NULL;
    p->pRight = NULL;
    return p;
}

void insertNode(BTNode *pNode, int value)
{
    assert(pNode);
    if( value<pNode->value )
    {
        if(pNode->pLeft)
        {
            insertNode(pNode->pLeft, value);
        }
        else
        {
            pNode->pLeft = newNode(value);
        }
    }
    else
    {
        if(pNode->pRight)
        {
            insertNode(pNode->pRight, value);
        }
        else
        {
            pNode->pRight = newNode(value);
        }
    }
}

void PrintNode(BTNode *pNode)
{
    if(pNode)
    {
        printf("%d ",pNode->value);
        PrintNode(pNode->pLeft);
        PrintNode(pNode->pRight);
    }
}

List* addListNode(BTNode *pNode)
{
    List *p = (List*)malloc(sizeof(List));
    p->pNode = pNode;
    p->pNext = NULL;
    return p;
}
void PrintByLine(BTNode *pNode)
{
    assert(pNode);
    List *pList1;
    List *pList2;
    List *p;
    List *pIndex;

    pList1 = (List*)malloc(sizeof(List));
    pList2 = (List*)malloc(sizeof(List));

    pList1->pNode = pNode;
    pList1->pNext = NULL;
    pList2->pNext = NULL;

    while(1)
    {

        pIndex = pList2;
        //遍历上一层结点，并输出
        for(p=pList1; p!=NULL; p=p->pNext)
        {
            printf("%d ",p->pNode->value);
            //输出的时候把下一层结点添加到pList2
            if( p->pNode->pLeft!=NULL )
            {
                pIndex->pNext = addListNode(p->pNode->pLeft);
                pIndex = pIndex->pNext;
            }
            if( p->pNode->pRight!=NULL )
            {
                pIndex->pNext = addListNode(p->pNode->pRight);
                pIndex = pIndex->pNext;
            }
        }
        //把上一层更新为下一层结点
        pList1 = pList2->pNext;
        pList2->pNext = NULL;
        if(!pList1)
        {
            break;
        }
    }
}


void bt_test(void)
{
    int a[]= {4, 47, 9, 1,
             2, 10, 11, 33,
             31,8 , 10, 17,
             3, 4 , 56 ,55};
    int i;
    BTNode *pHead = newNode(a[0]);
    for(i=1; i<sizeof(a)/sizeof(a[0]); i++)
    {
        insertNode(pHead, a[i]);
    }
    //PrintNode(pHead);
    PrintByLine(pHead);
}

2 两柱香问题

有2柱香，质量分布不均匀，每柱香烧完要1小时，怎么确定15分钟的时间。

不会，直接问百度，关键的是用其中一炷香点2头，另外一炷香点一头，等其中一柱烧完后，另一柱还剩半小时，然后再点燃另一头，烧完就是15分钟。