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);
}
}
DFS iterative in Java:
//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;
}
If you have pointers to parent nodes, you can do it without additional memory.
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
Note that if the child nodes are stored as an array rather than through sibling pointers, the next sibling can be found as:
def next_sibling(node):
try:
i = node.parent.child_nodes.index(node)
return node.parent.child_nodes[i+1]
except (IndexError, AttributeError):
return None
FULL example WORKING code, without stack:
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);
}}
output: Breadth First Traversal- starting from vertex 2: 2 0 3 1 4 Depth First Traversal- starting from vertex 2: 2 3 4 1 0
You can use a stack. I implemented graphs with Adjacency Matrix:
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();
}
}
}
http://www.youtube.com/watch?v=zLZhSSXAwxI
Just watched this video and came out with implementation. It looks easy for me to understand. Please critique this.
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()
}
Just wanted to add my python implementation to the long list of solutions. This non-recursive algorithm has discovery and finished events.
worklist = [root_node]
visited = set()
while worklist:
node = worklist[-1]
if node in visited:
# Node is finished
worklist.pop()
else:
# Node is discovered
visited.add(node)
for child in node.children:
worklist.append(child)
Use a stack to track your nodes
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)
}
An ES6 implementation based on biziclops great answer:
root = {_x000D_
text: "root",_x000D_
children: [{_x000D_
text: "c1",_x000D_
children: [{_x000D_
text: "c11"_x000D_
}, {_x000D_
text: "c12"_x000D_
}]_x000D_
}, {_x000D_
text: "c2",_x000D_
children: [{_x000D_
text: "c21"_x000D_
}, {_x000D_
text: "c22"_x000D_
}]_x000D_
}, ]_x000D_
}_x000D_
_x000D_
console.log("DFS:")_x000D_
DFS(root, node => node.children, node => console.log(node.text));_x000D_
_x000D_
console.log("BFS:")_x000D_
BFS(root, node => node.children, node => console.log(node.text));_x000D_
_x000D_
function BFS(root, getChildren, visit) {_x000D_
let nodesToVisit = [root];_x000D_
while (nodesToVisit.length > 0) {_x000D_
const currentNode = nodesToVisit.shift();_x000D_
nodesToVisit = [_x000D_
...nodesToVisit,_x000D_
...(getChildren(currentNode) || []),_x000D_
];_x000D_
visit(currentNode);_x000D_
}_x000D_
}_x000D_
_x000D_
function DFS(root, getChildren, visit) {_x000D_
let nodesToVisit = [root];_x000D_
while (nodesToVisit.length > 0) {_x000D_
const currentNode = nodesToVisit.shift();_x000D_
nodesToVisit = [_x000D_
...(getChildren(currentNode) || []),_x000D_
...nodesToVisit,_x000D_
];_x000D_
visit(currentNode);_x000D_
}_x000D_
}
_x000D_
You would use a stack that holds the nodes that were not visited yet:
stack.push(root)
while !stack.isEmpty() do
node = stack.pop()
for each node.childNodes do
stack.push(stack)
endfor
// …
endwhile
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);
}
}
The general logic is, push a node(starting from root) into the Stack, Pop() it and Print() value. Then if it has children( left and right) push them into the stack - push Right first so that you will visit Left child first(after visiting node itself). When stack is empty() you will have visited all nodes in Pre-Order.
Suppose you want to execute a notification when each node in a graph is visited. The simple recursive implementation is:
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
}
Ok, now you want a stack-based implementation because your example doesn't work. Complex graphs might for instance cause this to blow the stack of your program and you need to implement a non-recursive version. The biggest issue is to know when to issue a notification.
The following pseudo-code works (mix of Java and C++ for readability):
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);
}
}
It looks complicated but the extra logic needed for issuing notifications exists because you need to notify in reverse order of visit - DFS starts at root but notifies it last, unlike BFS which is very simple to implement.
For kicks, try following graph: nodes are s, t, v and w. directed edges are: s->t, s->v, t->w, v->w, and v->t. Run your own implementation of DFS and the order in which nodes should be visited must be: w, t, v, s A clumsy implementation of DFS would maybe notify t first and that indicates a bug. A recursive implementation of DFS would always reach w last.
Pseudo-code based on @biziclop's answer:
getNode(id)
and getChildren(id)
N
NOTE: I use array-indexing from 1, not 0.
Breadth-first
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
Depth-first
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
Here is a link to a java program showing DFS following both reccursive and non-reccursive methods and also calculating discovery and finish time, but no edge laleling.
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);
}
}
}
}
}
Full source here.
Non-recursive DFS using ES6 generators
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
}
}
It deviates from typical non-recursive DFS to easily detect when all reachable descendants of given node were processed and to maintain the current path in the list/stack.
Using Stack
, here are the steps to follow: Push the first vertex on the stack then,
Here's the Java program following the above steps:
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;
}
While "use a stack" might work as the answer to contrived interview question, in reality, it's just doing explicitly what a recursive program does behind the scenes.
Recursion uses the programs built-in stack. When you call a function, it pushes the arguments to the function onto the stack and when the function returns it does so by popping the program stack.
Source: Stackoverflow.com