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Posts tagged as “BFS”

花花酱 LeetCode 675. Cut Off Trees for Golf Event

https://leetcode.com/problems/cut-off-trees-for-golf-event/

Problem:

You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:

  1. 0 represents the obstacle can’t be reached.
  2. 1 represents the ground can be walked through.
  3. The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the tree’s height.

You are asked to cut off all the trees in this forest in the order of tree’s height – always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).

You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can’t cut off all the trees, output -1 in that situation.

You are guaranteed that no two trees have the same height and there is at least one tree needs to be cut off.

Example 1:

Example 2:

Example 3:

Hint: size of the given matrix will not exceed 50×50.

 

Idea:

Greedy + Shortest path

Identify and sort the trees by its heights, then find shortest paths between

0,0 to tree[1]
tree[1] to tree[2]

tree[n-1] to tree[n]

Time complexity: O(m^2n^2)

Space complexity: O(mn)

 

Solution:

 

花花酱 LeetCode 637. Average of Levels in Binary Tree

 

Problem:

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

 

Time Complexity:

O(n)

Space Complexity:

O(h)

Solution 1:

BFS

Solution 2:

DFS

 

Related Problems:

花花酱 LeetCode 102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its level order traversal as:

Solution 1: BFS O(n)

Solution 2: DFS O(n)