# Question

Formatted question description: https://leetcode.ca/all/286.html

 286	Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

-1     - A wall or an obstacle.
0      - A gate.
INF    - Infinity means an empty room.
We use the value 2^31 - 1 = 2147483647 to represent INF as you may assume that
the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate.
If it is impossible to reach a gate, it should be filled with INF.

For example, given the 2D grid:

INF    -1      0       INF
INF    INF     INF     -1
INF    -1      INF     -1
0      -1      INF     INF

After running your function, the 2D grid should be:

3  -1   0   1
2   2   1  -1
1  -1   2  -1
0  -1   3   4

@similar: 317	Shortest Distance from All Buildings


# Algorithm

Search for the position of 0, and every time a 0 is found, start DFS traversal with four adjacent points around it as the starting point, and bring in the depth value 1.

If the value encountered is greater than the current depth value, assign the position value to the current depth value, and start DFS traversal for the four adjacent points of the current point. Note that the depth value needs to be increased by 1.

After the traversal is completed, all positions are updated correctly

Java