DFS & BFS

DFS 深度优先

BFS 广度优先

DFS或者BFS都是在联通区域内遍历节点的方法

用在二叉树上DFS有preOreder，inOrder，postOrder，BFS就是层次遍历。

在二叉树上的节点，只有两个选择，left 和right，即，对于每一个节点，in 有1个， out 有两个，有向图

在矩阵的节点上，有四个选择，up、down、left和right四种选择，即，即，对于每一个节点，in 有4个， out 有4个，有向图

在surrounded Regions 中可以使用DFS或者BFS来解决问题

下面以4x4 矩阵为例，说明DFS和BFS的工作过程

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
using namespace std;

void printArray(int *array, int size)
{
    for(int i = 0; i < size; i++)
        cout << array[i]<< "	" ;
    cout << endl;
}

void printVector(vector<char> array )
{
    for(int i = 0; i <array.size(); i++)
        cout << array[i]<< "	" ;
    cout << endl;
}

void printVector(vector<int> array )
{
    for(int i = 0; i <array.size(); i++)
        cout << array[i]<< "	" ;
    cout << endl;
}

class Solution {
            queue<pair<int, int> > m_que;
    public:
        void dfs(vector<vector<int> > &board, int i, int j)
        {   
            size_t row = board.size();
            size_t col = board[0].size();

            if(i < 0 || i > row-1 || j < 0 || j > col-1)
                return;
            if( board[i][j] == INT_MAX)
                return;
                cout << "(" << i <<"," << j << ") = " <<  board[i][j] << endl ;
            board[i][j] = INT_MAX;//tag, in order to reverse back
            dfs(board, i, j-1);
            dfs(board, i, j+1);
            dfs(board, i-1, j);
            dfs(board, i+1, j);
        }

        void fill(vector<vector<int> > &board, int i, int j){
            size_t row = board.size();
            size_t col = board[0].size();

            if(i<0 || i>=row || j<0 || j>=col || board[i][j]== INT_MAX)
                return;

            pair<int, int> p ;
            p.first = i;
            p.second = j;
            m_que.push(p);

            cout << "(" << i <<"," << j << ") = " <<  board[i][j] << endl ;
            board[i][j]= INT_MAX;

        }


        void bfs(vector<vector<int> > &board, int i, int j)
        {
            fill(board, i, j);

            while(!m_que.empty())
            {
                pair<int, int> p = m_que.front() ;
                m_que.pop();

                i = p.first;
                j = p.second;

                fill(board, i, j-1);
                fill(board, i, j+1);
                fill(board, i-1, j);
                fill(board, i+1, j);
            }

        }

        void dfs(vector<vector<int> > board)
        {
            if (board.empty()) return;

            dfs(board, 2 ,2);

        }
        void bfs(vector<vector<int> > board)
        {
            if (board.empty()) return;

            bfs(board, 2 ,2);

        }
};

int main()
{
    vector<vector<int> > board;
    vector<int> a;
    a.resize(4, 0);

    a[0] = 0;
    a[1] = 1;
    a[2] = 2;
    a[3] = 3;
    board.push_back(a);

    a[0] = 4;
    a[1] = 5;
    a[2] = 6;
    a[3] = 7;
    board.push_back(a);

    a[0] = 8;
    a[1] = 9;
    a[2] = 10;
    a[3] = 11;
    board.push_back(a);

    a[0] = 12;
    a[1] = 13;
    a[2] = 14;
    a[3] = 15;
    board.push_back(a);


//    board.clear();
    Solution sl;


    for(int i = 0; i < board.size(); i++)
        printVector(board[i]);

    cout <<endl;
    cout << "dfs" <<endl;
    sl.dfs(board);
    cout << "bfs" <<endl;
    sl.bfs(board);

    for(int i = 0; i < board.size(); i++)
        printVector(board[i]);

    return 0;
}

打印结果

[root@localhost surroundedRegions]# ./a.out

矩阵
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15

dfs
(2,2) = 10
(2,1) = 9
(2,0) = 8
(1,0) = 4
(1,1) = 5
(1,2) = 6
(1,3) = 7
(0,3) = 3
(0,2) = 2
(0,1) = 1
(0,0) = 0
(2,3) = 11
(3,3) = 15
(3,2) = 14
(3,1) = 13
(3,0) = 12

bfs
(2,2) = 10
(2,1) = 9
(2,3) = 11
(1,2) = 6
(3,2) = 14
(2,0) = 8
(1,1) = 5
(3,1) = 13
(1,3) = 7
(3,3) = 15
(0,2) = 2
(1,0) = 4
(3,0) = 12
(0,1) = 1
(0,3) = 3
(0,0) = 0