BFS與DFS簡介與LC實例
作者：Bluemapleman([email protected])
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文章目錄
前言：BFS（Breath-First Search，廣度優先搜索）與DFS（Depth-First Search，深度優先搜索）是兩種針對樹與圖數據結構的遍歷或搜索算法，在樹與圖相關算法的考察中是非常常見的兩種解題思路。
文章目錄
Definition of DFS and BFS
DFS的wikipedia定義:
Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. The algorithm starts at the root node (selecting some arbitrary node as the root node in the case of a graph) and explores as far as possible along each branch before backtracking.
BFS的wikipedia定義:
Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a ‘search key’[1]), and explores all of the neighbor nodes at the present depth prior to moving on to the nodes at the next depth level.
It uses the opposite strategy as depth-first search, which instead explores the highest-depth nodes first before being forced to backtrack and expand shallower nodes.
So obviously, as their name suggest, DFS focuses on ‘depth’ when searching or traversing while BFS focuses on ‘breath’.
By the way, because of DFS’s feature, it’s easy to relate it with ‘Backtracking’ algorithm as the wiki definition mentions. The relationship between DFS and backtracking is well explained by Reed Copsey on StackOverflow:
Backtracking is a more general purpose algorithm.
Depth-First search is a specific form of backtracking related to searching tree structures. From Wikipedia:
One starts at the root (selecting some node as the root in the graph case) and explores as far as possible along each branch before backtracking.
It uses backtracking as part of its means of working with a tree, but is limited to a tree structure.
Backtracking, though, can be used on any type of structure where portions of the domain can be eliminated - whether or not it is a logical tree. The Wiki example uses a chessboard and a specific problem - you can look at a specific move, and eliminate it, then backtrack to the next possible move, eliminate it, etc.
How to Implement DFS and BFS
DFS
In tree structure, DFS means we always start from a root node and try to reach the leaf node as direct as possible before we have to backtrack.
Order in which the nodes are visited
In graph, DFS means we start from a random assigned node in the graph, and explores as far as possible along the branch before we have to backtrack.
So the key points for DFS are:
- How to explore as far as possible?
- How to backtrack?
How to explore as far as possible
Normally, for tree node, it would have left child or right child, so we would continuously go on exploring current node’s child node until we encounter a null node, then we go back to last node. Repeat above procedures until all nodes have been visited.
for graph node, we do the similar exploration: explore as further as possible according to the representation of graph (adjacency list, adjacency matrix or incidence matrix) until we find no more node that hasn’t been visited and connected with current node, then we go back to last node. Repeat above procedures until all nodes have been visited.
How to backtrack/go back?
‘Go back’ generally can be realized using data structure ——stack—— or by recursion. And if we use stack, it means we would need to push each node we visited in the process of exploring each branch, and pop when we can’t explore further starting from current node.
Time and Space Complexity
For time, every reachable vertex enters and leaves queue once. Every undirected edge is examined twice (Every directed edge is examined once). So total time is O(V+E).
Total Space: O(V) (Beyond space for G).
BFS
In tree structure, BFS means we always start from a root node and try to all the other nodes in the same breath before we further try exploring nodes at next depth level. (The same explanation for graph)
Order in which the nodes are visited
So the key points for BFS are:
- How to explore all nodes of same depth level?
How to explore all nodes of same depth level?
We can use a queue to do this: Starting from root node of a tree (Or a random node in a graph), we add all nodes connected with the starting node and add them to the queue. Then, we poll node from queue one by one and repeat above procedures until all nodes have been visited.
*Special properties of DFS
Classification of edges
We can define four edge types in terms of the depth-first forest G produced by a depth-first search on G:
1.Tree edges are edges in the depth-first forest G. Edge(u,v) is a tree edge if v was first discovered by exploring edge (u,v)
2.Back edges are those edges (u,v) connecting a vertex u to an ancestor v in a depth-first tree. We consider self-loops, which may occur in directed graphs, to be back edges.
3.Forward edges are those nontree edges (u,v) connecting a vertex u to a descendant v in a depth-first tree.
4.Cross edges are all other edges. They can go between vertices in the same
depth-first tree, as long as one vertex is not an ancestor of the other, or they can go between vertices in different depth-first trees.
Figure 22.5 Properties of depth-first search. (a) The result of a depth-first search of a directed graph. Vertices are timestamped and edge types are indicated as in Figure 22.4. (b) Intervals for the discovery time and finishing time of each vertex correspond to the parenthesization shown. Each
rectangle spans the interval given by the discovery and finishing times of the corresponding vertex. Only tree edges are shown. If two intervals overlap, then one is nested within the other, and the vertex corresponding to the smaller interval is a descendant of the vertex corresponding to the larger.
© The graph of part (a) redrawn with all tree and forward edges going down within a depth-first tree and all back edges going up from a descendant to an ancestor.
Topological Sort
A topological sort of a DAG(Directed Acyclic Graph) is a linear ordering of all its vertices such that if G contains an edge (u,v), then u appears before v in the ordering. (If the graph contains a cycle, then no linear ordering is possible.) We can view a topological sort of a graph as an ordering of its vertices along a horizontal line so that all directed edges go from left to right.
Corollary: In a DAG, reverse finishing order is a topological sort. (proving done in following picture)
- Example
Time and Space Complexity
Completely the same as BFS:
Time: O(V+E)
Space: O(V) (beyond space for G, including stacks if accomplished recursively)
Typical Leetcode Prbolems
DFS
Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]
- My Answer
package medium2;
import java.util.ArrayList;
import java.util.List;
/**
* @author Tom Qian
* @email [email protected]
* @github https://github.com/bluemapleman
* @date 2018年6月7日
*/
public class PathSumII
{
// DFS: make use of recursion to backtrack
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> ans=new ArrayList<List<Integer>>();
if(root==null)
return ans;
int goal=sum-root.val;
if(goal==0) {
if(root.left==null && root.right==null) {
List<Integer> tempList=new ArrayList<>();
tempList.add(root.val);
ans.add(tempList);
return ans;
}
}
List<List<Integer>> temp;
if((temp=pathSum(root.left, goal)).size()!=0) {
for(List<Integer> list:temp) {
list.add(0, root.val);
ans.add(list);
}
}
if((temp=pathSum(root.right, goal)).size()!=0) {
for(List<Integer> list:temp) {
list.add(0,root.val);
ans.add(list);
}
}
return ans;
}
}
Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
- My Answer
package medium2;
/**
* @author Tom Qian
* @email [email protected]
* @github https://github.com/bluemapleman
* @date 2018年6月11日
*/
public class ConvertSortedListtoBinarySearchTree
{
// DFS: make use of recursion to backtrack
// find the middle node of sorted linked list, and take it as the root node of the BST.
public TreeNode sortedListToBST(ListNode head) {
if(head==null)
return null;
ListNode slow=head,fast=head,followSlow=head;
boolean moveFlag=false;
while(fast!=null && fast.next!=null) {
if(moveFlag)
followSlow=followSlow.next;
moveFlag=true;
slow=slow.next;
fast=fast.next.next;
}
TreeNode root=new TreeNode(slow.val);
if(moveFlag) {
followSlow.next=null;
root.left=sortedListToBST(head);
root.right=sortedListToBST(slow.next);
}
return root;
}
}
Course Schedule
There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should
also have finished course 1. So it is impossible.
Note:
1.The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2.You may assume that there are no duplicate edges in the input prerequisites.
- My Answer
// DFS
public boolean canFinish(int numCourses, int[][] prerequisites) {
Map<Integer, List<Integer>> map=new HashMap<>();
for(int i=0;i<numCourses;i++)
map.put(i, new ArrayList<>());
for(int i=0;i<prerequisites.length;i++) {
map.get(prerequisites[i][0]).add(prerequisites[i][1]);
}
// start DFS: detect if there is any circle in course graph, i.e. whether DFS starting from certain start point i would lead to the start point again.
for(int i=0;i<numCourses;i++) {
// Use a set to avoid infinite loop: when met same node twice, ignore it.
Set<Integer> set=new HashSet<>();
// Use a stack to backtrack
ArrayDeque<Integer> stack=new ArrayDeque<>();
List<Integer> preCourseList=map.get(i);
for(Integer preCourse:preCourseList)
stack.push(preCourse);
while(!stack.isEmpty()) {
int preCourse=stack.pop();
if(set.contains(preCourse))
continue;
else
set.add(preCourse);
if(preCourse==i)
return false;
else {
preCourseList=map.get(preCourse);
for(Integer tempPreCourse:preCourseList) {
stack.push(tempPreCourse);
}
}
}
}
return true;
}
BFS
Course Schedule
- My Answer
// BFS
public boolean canFinish(int numCourses, int[][] prerequisites) {
Map<Integer, List<Integer>> map=new HashMap<>();
for(int i=0;i<numCourses;i++)
map.put(i, new ArrayList<>());
for(int i=0;i<prerequisites.length;i++) {
map.get(prerequisites[i][0]).add(prerequisites[i][1]);
}
// start DFS: detect if there is any circle in course graph, i.e. whether BFS starting from certain start point i would lead to the start point again.
for(int i=0;i<numCourses;i++) {
// Use a set to avoid infinite loop: when met same node twice, ignore it.
Set<Integer> set=new HashSet<>();
// Use a queue to remember nodes of same depth level
ArrayDeque<Integer> queue=new ArrayDeque<>();
List<Integer> preCourseList=map.get(i);
for(Integer preCourse:preCourseList)
queue.add(preCourse);
while(!queue.isEmpty()) {
int preCourse=queue.poll();
if(set.contains(preCourse))
continue;
else
set.add(preCourse);
if(preCourse==i)
return false;
else {
preCourseList=map.get(preCourse);
for(Integer tempPreCourse:preCourseList) {
queue.add(tempPreCourse);
}
}
}
}
return true;
}
Binary Tree Right Side View
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Example:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <---
/ \
2 3 <---
\ \
5 4 <---
- My Answer
package medium2;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
/**
* @author Tom Qian
* @email [email protected]
* @github https://github.com/bluemapleman
* @date 2018年6月12日
*/
public class BinaryTreeRightSideView
{
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ans=new ArrayList<>();
if(root==null)
return ans;
ans.add(root.val);
ArrayDeque<TreeNode> queue1=new ArrayDeque<>(),queue2=new ArrayDeque<>();;
queue1.add(root);
while(!queue1.isEmpty() || !queue2.isEmpty()){
TreeNode rightestNode=null;
if(!queue1.isEmpty()) {
while(!queue1.isEmpty()) {
TreeNode fatherNode=queue1.poll();
if(fatherNode.right!=null) {
queue2.add(fatherNode.right);
if(rightestNode==null)
rightestNode=fatherNode.right;
}
if(fatherNode.left!=null) {
queue2.add(fatherNode.left);
if(rightestNode==null)
rightestNode=fatherNode.left;
}
}
}else{
while(!queue2.isEmpty()) {
TreeNode fatherNode=queue2.poll();
if(fatherNode.right!=null) {
queue1.add(fatherNode.right);
if(rightestNode==null)
rightestNode=fatherNode.right;
}
if(fatherNode.left!=null) {
queue1.add(fatherNode.left);
if(rightestNode==null)
rightestNode=fatherNode.left;
}
}
}
if(rightestNode!=null)
ans.add(rightestNode.val);
}
return ans;
}
}