我正在寻找一种针对非二叉树的非递归深度优先搜索算法。很感谢任何形式的帮助。
我正在寻找一种针对非二叉树的非递归深度优先搜索算法。很感谢任何形式的帮助。
Answers:
DFS:
list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
currentnode = nodes_to_visit.take_first();
nodes_to_visit.prepend( currentnode.children );
//do something
}
BFS:
list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
currentnode = nodes_to_visit.take_first();
nodes_to_visit.append( currentnode.children );
//do something
}
两者的对称性很酷。
更新:如前所述,take_first()
删除并返回列表中的第一个元素。
.first()
功能还将元素从列表中删除。就像shift()
许多语言一样。pop()
也起作用,并以从右到左的顺序而不是从左到右的顺序返回子节点。
gray(1st)->gray(2nd)->gray(3rd)->blacken(3rd)->blacken(2nd)->blacken(1st)
。但是您的代码会产生:gray(1st)->gray(2nd)->gray(3rd)->blacken(2nd)->blacken(3rd)->blacken(1st)
。
您将使用一个堆栈来保存尚未访问的节点:
stack.push(root)
while !stack.isEmpty() do
node = stack.pop()
for each node.childNodes do
stack.push(stack)
endfor
// …
endwhile
if (nodes are not marked)
以判断是否适合将其推入堆栈。能行吗?
doing cycles
?我想我只想要DFS的命令。是吗,谢谢。
for each node.childNodes.reverse() do stack.push(stack) endfor
)推送子注释。这可能也是您想要的。这段视频为什么要这样解释:youtube.com/watch ?
如果您有指向父节点的指针,则无需额外的内存即可完成操作。
def dfs(root):
node = root
while True:
visit(node)
if node.first_child:
node = node.first_child # walk down
else:
while not node.next_sibling:
if node is root:
return
node = node.parent # walk up ...
node = node.next_sibling # ... and right
请注意,如果子节点存储为数组而不是通过同级指针存储,则下一个同级可以找到:
def next_sibling(node):
try:
i = node.parent.child_nodes.index(node)
return node.parent.child_nodes[i+1]
except (IndexError, AttributeError):
return None
while not node.next_sibling or node is root:
。
使用堆栈来跟踪您的节点
Stack<Node> s;
s.prepend(tree.head);
while(!s.empty) {
Node n = s.poll_front // gets first node
// do something with q?
for each child of n: s.prepend(child)
}
尽管“使用堆栈” 可以作为人为设计的面试问题的答案,但实际上,它只是在明确地执行递归程序在后台执行的操作。
递归使用程序内置堆栈。当您调用一个函数时,它将参数传递给该函数到堆栈上,而当函数返回时,它会通过弹出程序堆栈来执行。
基于biziclops的ES6实现很好的答案:
root = {
text: "root",
children: [{
text: "c1",
children: [{
text: "c11"
}, {
text: "c12"
}]
}, {
text: "c2",
children: [{
text: "c21"
}, {
text: "c22"
}]
}, ]
}
console.log("DFS:")
DFS(root, node => node.children, node => console.log(node.text));
console.log("BFS:")
BFS(root, node => node.children, node => console.log(node.text));
function BFS(root, getChildren, visit) {
let nodesToVisit = [root];
while (nodesToVisit.length > 0) {
const currentNode = nodesToVisit.shift();
nodesToVisit = [
...nodesToVisit,
...(getChildren(currentNode) || []),
];
visit(currentNode);
}
}
function DFS(root, getChildren, visit) {
let nodesToVisit = [root];
while (nodesToVisit.length > 0) {
const currentNode = nodesToVisit.shift();
nodesToVisit = [
...(getChildren(currentNode) || []),
...nodesToVisit,
];
visit(currentNode);
}
}
PreOrderTraversal is same as DFS in binary tree. You can do the same recursion
taking care of Stack as below.
public void IterativePreOrder(Tree root)
{
if (root == null)
return;
Stack s<Tree> = new Stack<Tree>();
s.Push(root);
while (s.Count != 0)
{
Tree b = s.Pop();
Console.Write(b.Data + " ");
if (b.Right != null)
s.Push(b.Right);
if (b.Left != null)
s.Push(b.Left);
}
}
一般逻辑是,将节点(从根开始)推入堆栈,将其Pop()并打印()值。然后,如果它有子级(左和右),则将它们推入堆栈-首先推入Right,这样您将首先访问Left子级(在访问节点本身之后)。当stack为empty()时,您将访问Pre-Order中的所有节点。
使用ES6生成器的非递归DFS
class Node {
constructor(name, childNodes) {
this.name = name;
this.childNodes = childNodes;
this.visited = false;
}
}
function *dfs(s) {
let stack = [];
stack.push(s);
stackLoop: while (stack.length) {
let u = stack[stack.length - 1]; // peek
if (!u.visited) {
u.visited = true; // grey - visited
yield u;
}
for (let v of u.childNodes) {
if (!v.visited) {
stack.push(v);
continue stackLoop;
}
}
stack.pop(); // black - all reachable descendants were processed
}
}
它不同于典型的非递归DFS,可以轻松检测给定节点的所有可到达后代何时被处理,并在列表/堆栈中维护当前路径。
假设您要在访问图形中的每个节点时执行通知。简单的递归实现是:
void DFSRecursive(Node n, Set<Node> visited) {
visited.add(n);
for (Node x : neighbors_of(n)) { // iterate over all neighbors
if (!visited.contains(x)) {
DFSRecursive(x, visited);
}
}
OnVisit(n); // callback to say node is finally visited, after all its non-visited neighbors
}
好的,现在您想要一个基于堆栈的实现,因为您的示例不起作用。例如,复杂的图形可能会导致这破坏程序堆栈,并且您需要实现非递归版本。最大的问题是知道何时发出通知。
以下伪代码有效(为了提高可读性,混合使用Java和C ++):
void DFS(Node root) {
Set<Node> visited;
Set<Node> toNotify; // nodes we want to notify
Stack<Node> stack;
stack.add(root);
toNotify.add(root); // we won't pop nodes from this until DFS is done
while (!stack.empty()) {
Node current = stack.pop();
visited.add(current);
for (Node x : neighbors_of(current)) {
if (!visited.contains(x)) {
stack.add(x);
toNotify.add(x);
}
}
}
// Now issue notifications. toNotifyStack might contain duplicates (will never
// happen in a tree but easily happens in a graph)
Set<Node> notified;
while (!toNotify.empty()) {
Node n = toNotify.pop();
if (!toNotify.contains(n)) {
OnVisit(n); // issue callback
toNotify.add(n);
}
}
它看起来很复杂,但是存在发出通知所需的额外逻辑,因为您需要以相反的访问顺序进行通知-DFS从根开始但最后通知它,这与BFS实施起来非常简单不同。
对于踢,请尝试以下图形:节点为s,t,v和w。有向边是:s-> t,s-> v,t-> w,v-> w和v-> t。运行您自己的DFS实现,访问节点的顺序必须为:w,t,v,s DFS的笨拙实现可能会首先通知t并表明存在错误。DFS的递归实现将始终达到最后。
完整示例工作代码,无堆栈:
import java.util.*;
class Graph {
private List<List<Integer>> adj;
Graph(int numOfVertices) {
this.adj = new ArrayList<>();
for (int i = 0; i < numOfVertices; ++i)
adj.add(i, new ArrayList<>());
}
void addEdge(int v, int w) {
adj.get(v).add(w); // Add w to v's list.
}
void DFS(int v) {
int nodesToVisitIndex = 0;
List<Integer> nodesToVisit = new ArrayList<>();
nodesToVisit.add(v);
while (nodesToVisitIndex < nodesToVisit.size()) {
Integer nextChild= nodesToVisit.get(nodesToVisitIndex++);// get the node and mark it as visited node by inc the index over the element.
for (Integer s : adj.get(nextChild)) {
if (!nodesToVisit.contains(s)) {
nodesToVisit.add(nodesToVisitIndex, s);// add the node to the HEAD of the unvisited nodes list.
}
}
System.out.println(nextChild);
}
}
void BFS(int v) {
int nodesToVisitIndex = 0;
List<Integer> nodesToVisit = new ArrayList<>();
nodesToVisit.add(v);
while (nodesToVisitIndex < nodesToVisit.size()) {
Integer nextChild= nodesToVisit.get(nodesToVisitIndex++);// get the node and mark it as visited node by inc the index over the element.
for (Integer s : adj.get(nextChild)) {
if (!nodesToVisit.contains(s)) {
nodesToVisit.add(s);// add the node to the END of the unvisited node list.
}
}
System.out.println(nextChild);
}
}
public static void main(String args[]) {
Graph g = new Graph(5);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
g.addEdge(3, 1);
g.addEdge(3, 4);
System.out.println("Breadth First Traversal- starting from vertex 2:");
g.BFS(2);
System.out.println("Depth First Traversal- starting from vertex 2:");
g.DFS(2);
}}
输出:广度优先遍历-从顶点2开始:2 0 3 1 4深度优先遍历-从顶点2:2 3 4 1 0
您可以使用堆栈。我用邻接矩阵实现了图:
void DFS(int current){
for(int i=1; i<N; i++) visit_table[i]=false;
myStack.push(current);
cout << current << " ";
while(!myStack.empty()){
current = myStack.top();
for(int i=0; i<N; i++){
if(AdjMatrix[current][i] == 1){
if(visit_table[i] == false){
myStack.push(i);
visit_table[i] = true;
cout << i << " ";
}
break;
}
else if(!myStack.empty())
myStack.pop();
}
}
}
Java中的DFS迭代:
//DFS: Iterative
private Boolean DFSIterative(Node root, int target) {
if (root == null)
return false;
Stack<Node> _stack = new Stack<Node>();
_stack.push(root);
while (_stack.size() > 0) {
Node temp = _stack.peek();
if (temp.data == target)
return true;
if (temp.left != null)
_stack.push(temp.left);
else if (temp.right != null)
_stack.push(temp.right);
else
_stack.pop();
}
return false;
}
http://www.youtube.com/watch?v=zLZhSSXAwxI
刚刚观看了此视频并提出了实施方案。这看起来很容易理解。请对此进行批评。
visited_node={root}
stack.push(root)
while(!stack.empty){
unvisited_node = get_unvisited_adj_nodes(stack.top());
If (unvisited_node!=null){
stack.push(unvisited_node);
visited_node+=unvisited_node;
}
else
stack.pop()
}
使用Stack
,执行以下步骤:将第一个顶点推入堆栈,然后,
这是按照上述步骤执行的Java程序:
public void searchDepthFirst() {
// begin at vertex 0
vertexList[0].wasVisited = true;
displayVertex(0);
stack.push(0);
while (!stack.isEmpty()) {
int adjacentVertex = getAdjacentUnvisitedVertex(stack.peek());
// if no such vertex
if (adjacentVertex == -1) {
stack.pop();
} else {
vertexList[adjacentVertex].wasVisited = true;
// Do something
stack.push(adjacentVertex);
}
}
// stack is empty, so we're done, reset flags
for (int j = 0; j < nVerts; j++)
vertexList[j].wasVisited = false;
}
Stack<Node> stack = new Stack<>();
stack.add(root);
while (!stack.isEmpty()) {
Node node = stack.pop();
System.out.print(node.getData() + " ");
Node right = node.getRight();
if (right != null) {
stack.push(right);
}
Node left = node.getLeft();
if (left != null) {
stack.push(left);
}
}
基于@biziclop答案的伪代码:
getNode(id)
和getChildren(id)
N
注意:我使用数组索引从1,而不是0。
广度优先
S = Array(N)
S[1] = 1; // root id
cur = 1;
last = 1
while cur <= last
id = S[cur]
node = getNode(id)
children = getChildren(id)
n = length(children)
for i = 1..n
S[ last+i ] = children[i]
end
last = last+n
cur = cur+1
visit(node)
end
深度优先
S = Array(N)
S[1] = 1; // root id
cur = 1;
while cur > 0
id = S[cur]
node = getNode(id)
children = getChildren(id)
n = length(children)
for i = 1..n
// assuming children are given left-to-right
S[ cur+i-1 ] = children[ n-i+1 ]
// otherwise
// S[ cur+i-1 ] = children[i]
end
cur = cur+n-1
visit(node)
end
这是一个指向Java程序的链接,该程序显示了DFS,它遵循递归和非递归方法,还计算发现和完成时间,但不进行边缘处理。
public void DFSIterative() {
Reset();
Stack<Vertex> s = new Stack<>();
for (Vertex v : vertices.values()) {
if (!v.visited) {
v.d = ++time;
v.visited = true;
s.push(v);
while (!s.isEmpty()) {
Vertex u = s.peek();
s.pop();
boolean bFinished = true;
for (Vertex w : u.adj) {
if (!w.visited) {
w.visited = true;
w.d = ++time;
w.p = u;
s.push(w);
bFinished = false;
break;
}
}
if (bFinished) {
u.f = ++time;
if (u.p != null)
s.push(u.p);
}
}
}
}
}
完整的源代码在这里。