面试金典--4.4

题目描述：给定二叉树，同一深度的放置在同一链表中，深度为D则有D个链表，BFS即可。

#include <iostream>
#include <queue>
#include <climits>
#include <algorithm>
#include <memory.h>
#include <stdio.h>
#include <ostream>
#include <vector>
#include <list>
#include <cmath>
#include <string>
#include <stdexcept>
#include <stack>
using namespace std;

typedef struct treeNode
{
    struct treeNode *left;
    struct treeNode *right;
    int val;
    treeNode(int pval):left(NULL),right(NULL),val(pval)
    {

    }
}Node;

list<list<Node*> > ans;
list<Node*> cur;
list<Node*> parent;
void fun(Node *root)
{
    if(root != NULL)
    {
        //cur = new list<Node*>();
        cur.push_back(root);
        while(cur.size() > 0)
        {
            ans.push_back(cur);
            parent = cur;
            //cur = new list
            cur.clear();
            Node* tmp;
            list<Node *>::iterator it = parent.begin();
            for( ; it != parent.end() ; ++it)
            {
                tmp = *it;
                cout<<tmp->val<<endl;
                if(tmp->left != NULL)
                    cur.push_back(tmp->left);
                if(tmp-> right != NULL)
                    cur.push_back(tmp->right);
            }
        }
    }
    return;
}

int main()
{
    Node tmp(1);
    Node *root = &tmp;
    Node tmp2(2);
    root->left = &tmp2;

    fun(root);
    list<list<Node *> >::iterator it = ans.begin();
    for( ; it != ans.end() ; ++it)
    {
        list<Node*> tmplis = *it;
        list<Node *>::iterator it1 = tmplis.begin();
        for( ; it1 != tmplis.end() ; ++it1)
        {
            Node * tmp3 = *it1;
            cout<<tmp3->val<<endl;
        }
    }
    return 0;
}